Contents
CHAPTER 4
4.1
4.2
4.3
4.4
CHAPTER
5
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
CHAPTER
6
6.1
6.2
6.3
6.4
6.5
6.6
6.7
CHAPTER 7
7.1
7.2
7.3
7.4
7.5
CHAPTER 8
8.1
8.2
8.3
8.4
8.5
8.6
The Chain Rule
Derivatives by the Chain Rule
Implicit Differentiation and Related Rates
Inverse Functions and Their Derivatives
Inverses of Trigonometric Functions
Integrals
The Idea of the Integral
Antiderivatives
Summation vs. Integration
Indefinite Integrals and Substitutions
The Definite Integral
Properties of the Integral and the Average Value
The Fundamental Theorem and Its Consequences
Numerical Integration
177
182
187
195
201
206
213
220
Exponentials and Logarithms
An Overview
The Exponential ex
Growth and Decay in Science and Economics
Logarithms
Separable Equations Including the Logistic Equation
Powers Instead of Exponentials
Hyperbolic Functions
Techniques of Integration
Integration by Parts
Trigonometric Integrals
Trigonometric Substitutions
Partial Fractions
Improper Integrals
Applications of the Integral
Areas and Volumes by Slices
Length of a Plane Curve
Area of a Surface of Revolution
Probability and Calculus
Masses and Moments
Force, Work, and Energy
228
236
242
252
259
267
277
CHAPTER
Integrals 5.1 The Idea of the Integral
This chapter is about the idea of integration, and also about the technique of integration. We explain how it is done in principle, and then how it is done in practice.
Integration is a problem of adding up infinitely many things, each of which is infinitesimally small. Doing the addition is not recommended. The whole point of calculus
is to offer a better way.
The problem of integration is to find a limit of sums. The key is to work backward
from a limit of differences (which is the derivative). We can integrate v(x) ifit turns
up as the derivative of another function f(x). The integral of v = cos x is f = sin x. The
integral of v = x is f = $x2. Basically, f(x) is an "antiderivative". The list of j ' s will
grow much longer (Section 5.4 is crucial). A selection is inside the cover of this book.
If we don't find a suitablef(x), numerical integration can still give an excellent answer.
I could go directly to the formulas for integrals, which allow you to compute areas
under the most amazing curves. (Area is the clearest example of adding up infinitely
many infinitely thin rectangles, so it always comes first. It is certainly not the only
problem that integral calculus can solve.) But I am really unwilling just to write down
formulas, and skip over all the ideas. Newton and Leibniz had an absolutely brilliant
intuition, and there is no reason why we can't share it.
They started with something simple. We will do the same.
SUMS A N D DIFFERENCES
Integrals and derivatives can be mostly explained by working (very briefly) with sums
and differences. Instead of functions, we have n ordinary numbers. The key idea is
nothing more than a basic fact of algebra. In the limit as n + co,it becomes the basic
fact of calculus. The step of "going to the limit" is the essential difference between
algebra and calculus! It has to be taken, in order to add up infinitely many
infinitesimals-but we start out this side of it.
To see what happens before the limiting step, we need two sets of n numbers. The
first set will be v,, v,, ..., v,, where suggests velocity. The second set of numbers
will be f, ,f,, ... ,f,, where f recalls the idea of distance. You might think d would
be a better symbol for distance, but that is needed for the dx and dy of calculus.
5 Integrals
A first example has n = 4:
f1,f2,f3,f4= 1, 3, 6, 10.
01, 212, v3, v 4 = L 2 , 3 , 4
The relation between the v's and f's is seen in that example. When you are given
1, 3, 6, 10, how do you produce 1, 2, 3, 4? By taking drerences. The difference
between 10 and 6 is 4. Subtracting 6 - 3 is 3. The differencef2 -fl = 3 - 1 is v2 = 2.
Each v is the difference between two f 's:
vj is the dierencefi
.
This is the discrete form of the derivative. I admit to a small difficulty at j = 1, from
the fact that there is no fo. The first v should be fl -fo, and the natural idea is to
agree that fo is zero. This need for a starting point will come back to haunt us (or
help us) in calculus.
Now look again at those same numbers-but start with v. From v = 1,2,3,4 how
do you produce f = 1,3,6, lo? By taking sums. The first two v's add to 3, which is f2.
The first three v's add to f3 = 6. The sum of all four v's is 1 + 2 + 3 + 4 = 10. Taking
sums is the opposite of taking di$erences.
That idea from algebra is the key to calculus. The sum& involves all the numbers
v, v2 + + vj. The difference vj involves only the two numbers f i - f i - . The fact
that one reverses the other is the "Fundamental Theorem." Calculus will change sums
to integrals and differences to derivatives-but why not let the key idea come through
now?
+
The differences of the f's add up to f, -fo . All f's in between are canceled, leaving
only the last fn and the starting foe The sum "telescopes":
+ U2 + 03 + ... + v n = (fl -fo) + (f2 -f1) + (f3 -f2) + ... + (fn -fn- 1)The number fl is canceled by -fl . Similarly -f2 cancels f2 and -f, cancels f3.
Eventually fn and -fo are left. When fo is zero, the sum is the finalf,.
That completes the algebra. We add the v's by finding the f 's.
01
Question How do you add the odd numbers 1 + 3 + 5 + -..+ 99 (the v's)?
Answer They are the differences between 0, 1,4,9, ... . Thesef's are squares. By the
Fundamental Theorem, the sum of 50 odd numbers is (50)2.
The tricky part is to discover the right f's! Their differences must produce the v's.
In calculus, the tricky part is to find the right f(x). Its derivative must produce v(x).
It is remarkable how often f can be found-more often for integrals than for sums.
Our next step is to understand how the integral is a limit of sums.
SUMS APPROACH INTEGRALS
Suppose you start a successful company. The rate of income is increasing. After
million dollars. In the first four years you reach
x years, the income per year is
$, $,and $ million dollars. Those numbers are displayed in a bar graph
(Figure S.la, for investors). I realize that most start-up companies make losses, but
your company is an exception. If the example is too good to be true, please keep
reading.
fi,
&
5.1 Thm Idea of the Integral
* Year
Fig. 5.1 Total income = total area of rectangles = 6.15.
fi,fi,fi.
The graph shows four rectangles, of heights ,,h,
Since the base of
each rectangle is one year, those numbers are also the areas of the rectangles. One
investor, possibly weak in arithmetic, asks a simple question: What is the total income
for all four years? There are two ways to answer, and I will give both.
+ + + $. Addition gives 6.15 million dollars.
The first answer is
Figure 5.lb shows this total-which is reached at year 4. This is exactly like velocities
and distances, but now v is the incomeper year andf is the totalincome.Algebraically,
vj.
fi is still v l
The second answer comes from geometry. The total income is the total area of the
rectangles. We are emphasizing the correspondence between athiition and area. That
point may seem obvious, but it becomes important when a second investor (smarter
than the first) asks a harder question.
f i fi f i
+ +
Here is the problem. The incomes as stated are false. The company did not make
a million dollars the first year. After three months, when x was 114, the rate of income
was only
= 112. The bar graph showed
= 1 for the whole year, but that was
an overstatement. The income in three months was not more than 112 times 114, the
. rate multiplied by the time.
All other quarters and years were also overstated. Figure 5.2a is closer to reality,
with 4 years divided into 16 quarters. It gives a new estimate for total income.
+ ,/16/4,
Again there are two ways to find the total. We add
remembering to multiply them all by 114 (because each rate applies to 114 year).
This is also the area of the 16 rectangles. The area approach is better because the 114
is automatic. Each rectangle has base 114, so that factor enters each area. The total
area is now 5.56 million dollars, closer to the truth.
You see what is coming. The next step divides time into weeks. After one week the
rate
is only J1/52. That is the height of the first rectangle-its base is Ax =
1/52. There is a rectangle for every week. Then a hard-working investor divides time
into days, and the base of each rectangle is Ax = 11365. At that point there are
4 x 365 = 1460 rectangles, or 1461 because of leap year, with a total area below 5)
&
fi
a+
fi
+
5
Integrals
Total income
= area of rectangles
1
= (sum of heights)
2.04 --
.768 -; " " " ,
..
0
0
..
0
0
I
I
I
I
I
I
I
I
I
I
I
I
I
I
I
Year
Fig. 5.2 Income = sum of areas (not heights)
million dollars. The calculation is elementary but depressing-adding up thousands
of square roots, each multiplied by A x from the base. There has to be a better way.
The better way, in fact the best way, is calculus. The whole idea is to allow for
continuous change. The geometry problem is to find the area under the square root
curve. That question cannot be answered by arithmetic, because it involves a limit.
The rectangles have base A x and heights &,
,,/%, ...
There are 4/Ax
rectangles-more and more terms from thinner and thinner rectangles. The area is
the limit of the sum as A x + 0.
This limiting area is the "integral." We are looking for a number below 54.
,d.
Algebra (area of n rectangles): Compute v, + + v, by finding f's.
Key idea: If vj =fj - f j , then the sum isf, -f,.
Calculus (area under curve): Compute the limit of Ax[v(Ax)+ v(2Ax) + ...I.
Key idea: If v(x)= dfldx then area = integral to be explained next.
. a -
- -
5.1
EXERCISES
Read-through questions
The problem of summation is to add v , + ... + v,. It is solved
if we find f ' s such that vj = a . Then v, + ... + v, equals
b
. The cancellation in ( f l -f,)
(f2 -f , )
...
(f,-,f, - ) leaves only c . Taking sums is the d of
taking differences.
,
+
The differences between 0, 1,4, 9 are v,, v,, o, =
For jj =j the difference between f l , and f, is v,, =
From this pattern 1 + 3 + 5 + ... + 19 equals g .
+ +
e
.
f
.
For functions, finding the integral is the reverse of h .
i
of v(x) is f(x).
If the derivative of f ( x ) is v(x), then the
If V ( X ) = l o x then f ( x ) = i . This is the k of a triangle
with base x and height lox.
Integrals begin with sums. The triangle under v = l o x out
to x = 4 has area I . It is approximated by four rectangles
of heights 10, 20, 30, 40 and area m . It is better approximated by eight rectangles of heights n and area o .
For n rectangles covering the triangle the area is the sum of
. As n -+ cc this sum should approach the number
P
. That is the integral of v = lOxfrom 0 to 4.
CI
181
5.1 The Idea of the Integral
Problems 1-6 are about sumsfj and differences vj.
1 With v = 1, 2, 4, 8, the formula for vj is
Find f, ,f 2 , f,, f, starting from fo = 0. What is f,?
(not 2j).
2 The same v = 1,2,4,8, . . . are the differences between
f = 1, 2, 4, 8, 16, .... Now fo = 1 and f j = 2j. (a) Check that
2 5 - 2 4 e q u a l ~v,. (b) What is 1 + 2 + 4 + 8 + l6?
3 The differences between f = 1, 112, 114, 118 are v =
negative v's do not add up to these
positive f's. Verify that u, + 11, + v, =f, -fo is still true.
- 112, - 114, - 118. These
4 Any constant C can be added to the antiderivative f(x)
because the
of a constant is zero. Any C can be
added to fo,f, , . . . because the
between the f's is
not changed.
5 Show thath = rj/(r - 1) hash -f,- = rj-'. Therefore the
geometric series 1 + r + .-. + rj-' adds up to
(remember to subtract f,).
'.
6 The sums h = (rj - l)/(r - 1) also have f j -fj-, = rjNow fo =
. Therefore 1 + r + ... + rj-' adds up to
f,.Thesuml+r+...+rnequals
.
7 Suppose v(x)= 3 for x < 1 and v(x) = 7 for x > 1. Find the
area f(x) from 0 to x, under the graph of v(x). (Two pieces.)
8 If v = 1, - 2, 3, -4, ..., write down the f's starting from
fo = 0. Find formulas for vj andfj when j is odd and j is even.
Problems 9-16 are about the company earning
& per year.
9 When time is divided into weeks there are 4 x 52 = 208
rectangles. Write down the first area, the 208th area, and the
jth area.
10 How do you know that the sum over 208 weeks is smaller
than the sum over 16 quarters?
&
11 A pessimist would use
at the beginning of each time
period as the income rate for that period. Redraw Figure 5.1
(both parts) using heights
,,h,,,.b, . How much lower
is the estimate of total income?
4,
4.
12 The same pessimist would redraw Figure 5.2 with heights
0,
. ... What is the height of the last rectangle? How
much does this change reduce the total rectangular area 5.56?
m,
13 At every step from years to weeks to days to hours, the
pessimist's area goes
and the optimist's area goes
. The difference between them is the area of the last
14 The optimist and pessimist arrive at the same limit as
years are divided into weeks, days, hours, seconds. Draw the
curve between the rectangles to show why the pessimist
is always too low and the optimist is too high.
&
fi
15 (Important) Let f(x) be the area under the
curve, above
the interval from 0 to x. The area to x + Ax is f(x + Ax). The
extra area is Af =
. This is almost a rectangle with
base
and height
So Af/Ax is close to
.
As Ax + 0 we suspect that dfldx =
.
&.
fi
16 Draw the
curve from x = 0 to 4 and put triangles
below to prove that the area under it is more than 5. Look
left and right from the point where
= 1.
fi
Problems 17-22 are about a company whose expense rate
v(x) = 6 - x is decreasing.
17 The expenses drop to zero at x =
expense during those years equals
of - -
. The total
. This is the area
18 The rectangles of heights 6, 5, 4, 3, 2, 1 give a total
estimated expense of
. Draw them enclosing the
triangle to show why this total is too high.
19 How many rectangles (enclosing the triangle) would you
need before their areas are within 1 of the correct triangular
area?
20 The accountant uses 2-year intervals and computes v =
5, 3, 1 at the midpoints (the odd-numbered years). What is
her estimate, how accurate is it, and why?
21 What is the area f(x) under the line v(x) = 6 - x above the
interval from 2 to x? What is the derivative of this f(x)?
22 What is the area f(x) under the line v(x) = 6 - x above the
interval from x to 6? What is the derivative of this f(x)?
23 With Ax = 113, find the area of the three rectangles that
enclose the graph of v(x)= x2.
fi
24 Draw graphs of v =
and v = x2 from 0 to 1. Which
.
areas add to l? The same is true for 11 = x3 and v =
25 From x to x +Ax, the area under v = x2 is AJ: This
. So
is almost a rectangle with base Ax and height
Af1A.u is close to
. In the limit we find dfldx = x2
and f(x) =
.
26 Compute the area of 208 rectangles under v(x) =
x=Otox=4.
& from
5.2
-
Antiderivatives The symbol was invented by Leibniz to represent the integral. It is a stretched-out
S , from the Latin word for sum. This symbol is a powerful reminder of the whole
and rectangular area
construction: Sum approaches integral, S approaches
approaches curved area:
I,
5fi
dx.
curved area = l v(x) dx =
(1)
The rectangles of base Ax lead to this limit-the integral of
The "dx" indicates
that Ax approaches zero. The heights vj of the rectangles are the heights v(x) of the
curve. The sum of vj times Ax approaches "the integral of v of x dx." You can imagine
an infinitely thin rectangle above every point, instead of ordinary rectangles above
special points.
We now find the area under the square root curve. The "limits of integration" are
0 and 4. The lower limit is x = 0, where the area begins. (The start could be any point
x = a.) The upper limit is x = 4, since we stop after four years. (The Jinish could be
any point x = b.) The area of the rectangles is a sum of base Ax times heights
The curved area is the limit of this sum. That l i d is the integral of &porn 0 to 4:
&.
&.
The outstanding problem of integral calculus is still to be solved. What is this limiting
area? We have a symbol for the answer, involving and
and dx-but we don't
have a number.
I
&
THE ANTIDERIVATIVE
&.
I wish I knew who discovered the area under the graph of
It may have been
Newton. The answer was available earlier, but the key idea was shared by Newton
and Leibniz. They understood the parallels between sums and integrals, and between
differences and derivatives. I can give the answer, by following that analogy. I can't
give the proof (yet)-it is the Fundamental Theorem of Calculus.
In algebra the differencef;. -f;.-, is vj. When we add, the sum of the v's isf. -fo.
In calculus the derivative of f(x) is v(x). When we integrate, the area under the v(x)
curve is f(x) minus f(0). Our problem asks for the area out to x = 4:
II
50 (Discrete vs. continuous, rectangles vs. curved areas, addition vs.
integration) laAe integral of 4 x ) ib the wnence iir fix):
rfdfldx =
fi then area =
dx =f(4) -fo.
(3)
I1 &,
&.
What is f(x)? Instead of the derivative of
we need its "antiderivative." We have
It is the opposite of Chapters 2-4, and
to find a functionf(x) whose derivative is
requires us to work backwards. The derivative of xn is nxn-'-now we need the
antiderivative. The quick formula is f(x) = xn+'/(n + 1)-we aim to understand it.
Solution Since the derivative lowers the exponent, the antiderivative raises it. We
go from x'I2 to x3I2. But then the derivative is (3/2)x1I2.It contains an unwanted
factor 312. To cancel that factor, put 213 into the antiderivative:
f(x) = 3x3I2has the required derivative V(X)= x 'I2
=
&.
2 43/2=16
Total income = 3
3
Rate of income =
a=
e
T1
I
-.
I
2
3
Year
,
I
c
2
1
3
: Year
4
Fig. 5.3 The integral of v(x) = &
, is the exact area 1613 under the curve.
There you see the key to integrals: Work backward from derivatives (and adjust).
Now comes a number-the exact area. At x = 4 we find x3I2= 8. Multiply by 213
to get 1613. Then subtract f(0) = 0:
The total income over four years is 1613 = 53 million dollars. This is f(4) -f(0). The
sum from thousands of rectangles was slowly approaching this exact area 5f.
= 3 million dollars.
Other areas The income in the first year, at x = 1, is
(The false income was 1 million dollars.) The total income after x years is 3x3I2,
which is the antiderivativef(x). The square root curve covers 213 of the overall rectangle
with area x3I2,and 213 of that
it sits in. The rectangle goes out to x and up to
rectangle is below the curve. (113 is above.)
&,
Other antiderivatives The derivative of x5 is 5x4. Therefore the antiderivative of x4
is x5/5. Divide by 5 (or n + 1) to cancel the 5 (or n + 1) from the derivative. And don't
allow n + 1 = 0:
The derivative v(x) = xn has the antiderivative f(x) = xn+' / ( n+ 1).
EXAMPLE 1 The antiderivative of x2 is ix3. This is the area under the parabola
v(x) = x2. The area out to x = 1 is
- f (0)3, or 113.
&
&
and x2 The 213 from
and the 113 from x2 add to 1. Those are
Remark on
curve, in the corner of Figure 5.3. If you turn the
the areas below and above the
and x = y2 are inverses!
curve by 90°, it becomes the parabola. The functions y =
The areas for these inverse functions add to a square of area 1.
&
&
AREA UNDER A STRAIGHT LINE
You already know the area of a triangle. The region is below the diagonal line v = x
in Figure 5.4. The base is 4, the height is 4, and the area is g4)(4) = 8. Integration is
5
Integrals
Exact area = 8
Area under v (x) = x
u (x) = S
Fig. 5.4 Triangular area 8 as the limit of rectangular areas 10, 9, 83, ....
not required! But if you allow calculus to repeat that answer, and build up the integral
f(x) = +x2 as the limiting area of many rectangles, you will have the beginning of
something important.
The four rectangles have area 1 + 2 + 3 + 4 = 10. That is greater than 8, because
the triangle is inside. 10 is a first approximation to the triangular area 8, and to
improve it we need more rectangles.
The next rectangles will be thinner, of width Ax = 112 instead of the original
Ax = 1. There will be eight rectangles instead of four. They extend above the line,
so the answer is still too high. The new heights are 112, 1, 312, 2, 512, 3, 712, 4. The
total area in Figure 5.4b is the sum of the base Ax = 112 times those heights:
area = $($ + 1 + $ + 2
+ + 4) = 9 (which is closer to 8).
Question What is the area of 16 rectangles? Their heights are $, 3, ... , 4.
Answer With base A x = $ the area is $($+++
+4)=8$.
The effort of doing the addition is increasing. A formula for the sums is needed, and
will be established soon. (The next answer would be 84.) But more important than
the formula is the idea. We are carrying out a Iimiting process, one step at a time. The
area of the rectangles is approaching the area of the triangle, as Ax decreases. The
same limiting process will apply to other areas, in which the region is much more
complicated. Therefore we pause to comment on what is important.
Area Under a Curve
What requirements are imposed on those thinner and thinner rectangles? It is not
essential that they all have the same width. And it is not required that they cover the
triangle completely. The rectangles could lie below the curve. The limiting answer
will still be 8, even if the widths Ax are unequal and the rectangles fit inside the
triangle or across it. We only impose two rules:
1. The largest width Ax,,, must approach zero.
2. The top of each rectangle must touch or cross the curve.
The area under the graph is defined to be the limit of these rectangular areas, if that
limit exists. For the straight line, the limit does exist and equals 8. That limit is
independent of the particular widths and heights-as we absolutely insist it should
be.
Section 5.5 allows any continuous v(x). The question will be the same-Does the
limit exist? The answer will be the same- Yes. That limit will be the integral of v(x),
and it will be the area under the curve. It will be f(x).
EXAMPLE 2 The triangular area from 0 to x is f(base)(height) = f(x)(x). That is
f(x) = f x2. Its derivative is v(x) = x. But notice that fx2 + 1 has the same derivative.
So does f = f x2 C, for any constant C. There is a "constant of integration" in f(x),
which is wiped out in its derivative v(x).
+
EXAMPLE 3 Suppose the velocity is decreasing: v(x) = 4 - x. If we sample v at x =
1,2,3,4, the rectangles lie under the graph. Because v is decreasing, the right end of
Then the rectangular area 3 + 2 + 1 + 0 = 6 is less than the
each interval gives v,,.
exact area 8. The rectangles are inside the triangle, and eight rectangles with base 4
come closer:
f 0) = 7.
rectangular area = f(3f + 3 +
+ +
Sixteen rectangles would have area 7f. We repeat that the rectangles need not have
the same widths Ax, but it makes these calculations easier.
What is the area out to an arbitrary point (like x = 3 or x = l)? We could insert
rectangles, but the Fundamental Theorem offers a faster way. Any antiderivative of
4 - x will give the area. We lookfor afunction whose derivative is 4 - x. The derivative
of 4x is 4, the derivative of fx2 is x, so work backward:
to achieve dfldx = 4 - x choose f(x) = 4x - f x2.
Calculus skips past the rectangles and computes f(3) = 7f. The area between x = 1
and x = 3 is the dference 77:- 3f = 4. In Figure 5.5, this is the area of the trapezoid.
The f-curve flattens out when the v-curve touches zero. No new area is being added.
1
2
3
Fig. 5.5 The area is Af
4
= 74 - 34 = 4.
1
2
3
4
Since v(x) decreases,f ( x ) bends down.
INDEFINITE INTEGRALS AND DEFINITE INTEGRALS
We have to distinguish two different kinds of integrals. They both use the antiderivative f(x). The definite one involves the limits 0 and 4, the indefinite one doesn't:
The indefinite integral is a function f(x) = 4x - i x 2 .
The definite integral from x = 0 to x = 4 is the number f(4) -f(0).
The definite integral is definitely 8. But the indefinite integral is not necessarily
4x - $x2. We can change f(x) by a constant without changing its derivative (since the
derivative of a constant is zero). The following functions are also antiderivatives:
The first two are particular examples. The last is the general case. The constant C
can be anything (including zero), to give all functions with the required derivative.
The theory of calculus will show that there are no others. The indefinite integral is
the most general antiderivative (with no limits):
indefinite integral f(x) = J v(x) dx = 4x - $ x 2 + C.
(5)
By contrast, the definite integral is a number. It contains no arbitrary constant C.
More that that, it contains no variable x. The definite integral is determined by the
function v(x) and the limits of integration (also known as the endpoints). It is the area
under the graph between those endpoints.
To see the relation of indefinite to definite, answer this question: What is the definite
integral between x = 1 and x = 3? The indefinite integral gives f(3) = 74 + C and
f(1) = 3f + C. To find the area between the limits, subtract f at one limit from f at the
other limit:
The constant cancels itself! The definite integral is the diflerence between the values
of the indefinite integral. C disappears in the subtraction.
The differencef(3) -f(l) is like fn -f,. The sum of v j from 1 to n has become "the
integral of v(x) from 1 to 3." Section 5.3 computes other areas from sums, and 5.4
computes many more from antiderivatives. Then we come back to the definite integral
and the Fundamental Theorem:
5.2 EXERCISES
Read-through questions
Integration yields the a under a curve y = v(x). It starts
from rectangles with base b and heights v(x) and areas
c
. As Ax -+ 0 the area v,Ax + + v,Ax becomes the
d
of ~ ( x )The
. symbol for the indefinite integral of v(x) is
The problem of integration is solved if we find f(x) such
that
f
. Then f is the g of v, and S:v(x) dx equals
h
minus i . The limits of integration are i . This
is a k integral, which is a I and not a function f(x).
Find an antiderivative f(x) for v(x) in 1-14. Then compute the
definite integral 1; u(x) dx =f(1) -f(0).
1 5x4 + 4 x 5
2 x + 12x2
3 I/& (or x - l")
4 (&)3
7 2 sin
+sin
8 sec2x + 1
zx
lo
COS
11 sin
(or x3I2)
cos
sin
(by experiment)
12 sin2x cos x
13 0 (find all f )
14 - 1 (find all f )
The example v(x) = x has f(x) = m . It also has f(x) =
" The area under from to is
The 'Onstant
is canceled in computing the difference P minus q .
If V(X)= x8 then f(x) = r .
v3
The sum v, + + v, =f, -fo leads to the Fundamental
t
integral is f(x) and
Theorem v(x) dx = s . The
the LJ integral is f(b) -f(a). Finding the v under the
v-graph is the opposite of finding the w of thef-graph.
16 The areas include a factor Ax, the base of each rectangle.
So the sum of v's is multiplied by
to approach the
integral. The difference of f's is divided by
to
approach the derivative.
z.
*
1:
15 If dfldx
is
= v(x) then
+ . - -+ u7 is
. If
the definite integral of v(x) from a to
= uj then the definite sum of
f, -fj-,
.
pv\; /
5.3 Summation versus Integration
17 The areas of 4, 8, and 16 rectangles were 10, 9, and 83,
containing the triangle out to x = 4. Find a formula for the
area AN of N rectangles and test it for N = 3 and N = 6.
18 Draw four rectangles with base 1 below the y = x line, and
find the total area. What is the area with N rectangles?
19 Draw y = sin x from 0 to 11. Three rectangles (base 11.13)
and six rectangles (base 11.16)contain an arch of the sine function. Find the areas and guess the limit.
20 Draw an example where three lower rectangles under a
curve (heights m,, m2, m3)have less area than two rectangles.
21 Draw y = l/x2 for 0 < x < 1 with two rectangles under it
(base 112). What is their area, and what is the area for four
rectangles? Guess the limit.
22 Repeat Problem 21 for y = llx.
23 (with calculator) For v(x) = I/& take enough rectangles
over 0 < x < 1 to convince any reasonable professor that the
area is 2. Find Ax) and verify that f(1) -f(0) = 2.
24 Find the area under the parabola v = x2 from x = 0 to
x = 4. Relate it to the area 1613 below
&.
25 For vl and v2 in the figure estimate the areasf(2) and f(4).
Start with f(0) = 0.
0
10
26 Draw y = v(x) so that the area Ax) increases until x = 1,
stays constant to x = 2, and decreases to f(3) = 1.
27 Describe the indefinite integrals of vl and u2. Do the areas
increase? Increase then decrease? ...
28 For v4(x)find the areaf(4) -f(1). Draw f4(x).
29 The graph of B(t) shows the birth rate: births per unit time
at time t. D(t) is the death rate. In what way do these numbers
appear on the graph?
1. The change in population from t = 0 to t = 10.
2. The time T when the population was largest.
3. The time t* when the population increased fastest.
30 Draw the graph of a function y4(x) whose area function
is v4(x).
31 If v2(x)is an antiderivative of y2(x), draw y2(x).
32 Suppose u(x) increases from 40) = 0 to v(3) = 4. The area
under y = v(x) plus the area on the left side of x = v-'(y)
.
equals
33 True or false, whenflx) is an antiderivative of u(x).
(a) 2f(x) is an antiderivative of 2v(x) (try examples)
(b) f(2x) is an antiderivative of v(2x)
(c) f(x) + 1 is an antiderivative of v(x) + 1
(d)f(x + 1) is an antiderivative of v(x + I).
(e) ( f ( ~ )is) ~an antiderivative of ( 4 ~ ) ) ~ .
5.3 Summation versus Integration
-
This section does integration the hard way. We find explicit formulas for f, =
u, + + u, . From areas of rectangles, the limits produce the area f(x) under a curve.
According to the Fundamental Theorem, dfldx should return us to v(x)-and we
verify in each case that it does.
May I recall that there is sometimes an easier way? If we can find an f(x) whose
derivative is u(x), then the integral of u is$ Sums and limits are not required, when f
is spotted directly. The next section, which explains how to look for f(x), will displace
this one. (If we can't find an antiderivative we fall back on summation.) Given a
successful f, adding any constant produces another f-since the derivative of the
constant is zero. The right constant achievesf(0) = 0, with no extra effort.
5
Integrals
This section constructs f(x) from sums. The next section searches for antiderivatives.
THE SIGMA NOTATION
In a section about sums, there has to be a decent way to express them. Consider
l 2 + 2' + 32 + 42. The individual terms are vj = j2. Their sum can be written in summation notation, using the capital Greek letter C (pronounced sigma):
1'
x
4
+ 2' + 32 + 42 is written
j2.
j=1
Spoken aloud, that becomes "the sum of j 2 from j = 1 to 4." It equals 30. The limits
on j (written below and above C) indicate where to start and stop:
The k at the end of ( 1 ) makes an additional point. There is nothing special about the
letter j. That is a "dummy variable," no better and no worse than k (or i). Dummy
variables are only on one side (the side with C),and they have no effect on the sum.
The upper limit n is on both sides. Here are six sums:
1
1 1
'f 7
=I +
+ + ... = 2 [infinite series]
2
2 4
-
-
k=O
The numbers 1 and n or 1 and 4 (or 0 and K ) are the lower limit and upper limit.
The dummy variable i or j or k is the index of summation. I hope it seems reasonable
that the infinite series 1 + 3 + $ + adds to 2. We will come back to it in Chapter 10.t
A sum like Z:=, 6 looks meaningless, but it is actually 6 + 6 + ... + 6 = 6n.
It follows the rules. In fact C:=, j 2 is not meaningless either. Every term is j 2 and
by the same rules. that sum is 4j2. However the i was probably intended to be j.
Then the sum is 1 + 4 + 9 + 16 = 30.
Question What happens to these sums when the upper limits are changed to n?
Answer The sum depends on the stopping point n. A formula is required (when
possible). Integrals stop at .u, sums stop at n, and we now look for special cases when
.f(.u) or *f,can be found.
A SPECIAL SUMMATION FORMULA
How do you add the first 100 whole numbers? The problem is to compute
tZeno the Greek believed it was impossible to get anywhere, since he would only go halfway
and then half again and half again. Infinite series would have changed his whole life.
5.3 Summation versus Integration
If you were Gauss, you would see the answer at once. (He solved this problem at a
ridiculous age, which gave his friends the idea of getting him into another class.) His
solution was to combine 1 + 100, and 2 + 99, and 3 + 98, always adding to 101. There
are fifty of those combinations. Thus the sum is (50)(101)= 5050.
The sum from 1 to n uses the same idea. The first and last terms add to n + 1. The
next terms n - 1 and 2 also add to n + 1. If n is even (as 100 was) then there are i n
parts. Therefore the sum is i n times n + 1:
The important term is i n 2 , but the exact sum is i n 2 + i n .
What happens if n is an odd number (like n = 99)? Formula (2) remains true. The
combinations 1 + 99 and 2 + 98 still add to n + 1 = 100. There are 399) = 493 such
pairs, because the middle term (which is 50) has nothing to combine with. Thus
1 + 2 + + 99 equals 493 times 100, or 4950.
Remark That sum had to be 4950, because it is 5050 minus 100. The sum up to 99
equals the sum up to 100 with the last term removed. Our key formula fn -fn- = v,
has turned up again!
EXAMPLE Find the sum 101
+ 102 + ... + 200 of the second hundred numbers.
First solution This is the sum from 1 to 200 minus the sum from 1 to 100:
The middle sum is $(200)(201) and the last is i(100)(101). Their difference is 15050.
Note! I left out '7 = "in the limits. It is there, but not written.
Second solution The answer 15050 is exactly the sum of the first hundred numbers
(which was 5050) plus an additional 10000. Believing that a number like 10000 can
never turn up by accident, we look for a reason. It is found through changing the
limits of summation:
200
(k + 100).
j is the same sum as
j = 101
k=l
This is important, to be able to shift limits around. Often the lower limit is moved
to zero or one, for convenience. Both sums have 100 terms (that doesn't change). The
dummy variable j is replaced by another dummy variable k. They are related by
j = k + 100 or equivalently by k =j - 100.
The variable must change everywhere-in the lower limit and the upper limit as
well as inside the sum. If j starts at 101, then k =j - 100 starts at 1. If j ends at 200,
k ends at 100. If j appears in the sum, it is replaced by k + 100 (and if j2 appeared it
would become (k +
From equation (4) you see why the answer is 15050. The sum 1 + 2 + ... + 100 is
5050 as before. 100 is added to each of those 100 terms. That gives 10000.
EXAMPLES OF CHANGING THE VARIABLE (and the limits)
3
4
i =0
j= 1
1 2' equals 1 2 ' '
1 viequals
i=3
..
(here i = j - 1). Both sums are 1 + 2 + 4 + 8
-
uj+,
j=O
(here i = j + 3 a n d j = i - 3 ) . Bothsums are v 3 + - . + v n .
Why change n to n - 3? Because the upper limit is i = n. So j + 3 = n and j = n - 3.
A final step is possible, and you will often see it. The new variable j can be changed
back to i. Dummy variables have no meaning of their own, but at first the result
looks surprising:
5
6
i =0
j=1
C 2' equals 2 2'- ' equals 2 zi- '.
i= 1
With practice you might do that in one step, skipping the temporary letter j. Every
i on the left becomes i - 1 on the right. Then i = 0, ..., 5 changes to i = 1, ..., 6 . (At
first two steps are safer.) This may seem a minor point, but soon we will be changing
the limits on integrals instead of sums. Integration is parallel to summation, and it
is better to see a "change of variable" here first.
+
Note about 1 2 + .-.+ n. The good thing is that Gauss found the sum f n(n + 1).
The bad thing is that his method looked too much like a trick. I would like to show
how this fits the fundamental rule connecting sums and differences:
Gauss says thatf, is f n(n + 1). Reducing n by 1, his formula for&-, is f (n - 1)n. The
dference f, - f,-, should be the last term n in the sum:
This is the one term v, = n that is included inf, but not inf,-I .
There is a deeper point here. For any sum f,, there are two things to check. The
f's must begin correctly and they must change correctly. The underlying idea is
mathematical induction: Assume the statement is true below n. Prove it for n.
Goat
To prove that 1 + 2 + --.+ n = f n(n + 1). This is the guess f,.
Proof by induction: Check fl (it equals 1). Check f, -f, - (it equals n).
For n = 1 the answer fn(n + 1) = f 1 2 is correct. For n = 2 this formula f 2 3
agrees with 1 + 2. But that separate test is not necessary! Iffl is right, and i f the
changef, -f,-, is right for every n, thenf, must be right. Equation (6) was the key
test, to show that the change in f's agrees with v.
That is the logic behind mathematical induction, but I am not happy with most
of the exercises that use it. There is absolutely no excitement. The answer is given by
some higher power (like Gauss), and it is proved correct by some lower power (like
us). It is much better when we lower powers find the answer for ourse1ves.t Therefore
I will try to do that for the second problem, which is the sum of squares.
THE SUM OF j2AND THE INTEGRAL OF x2
An important calculation comes next. It is the area in Figure 5.6. One region is made
up of rectangles, so its area is a sum of n pieces. The other region lies under the
parabola v = x2. It cannot be divided into rectangles, and calculus is needed.
The first problem is to find f, = 1' + 22 + 32 + + n2. This is a sum of squares,
with fl = 1 and f2 = 5 and f, = 14. The goal is to find the pattern in that sequence.
By trying to guessf, we are copying what will soon be done for integrals.
Calculus looks for an f(x) whose derivative is v(x). There f is an antiderivative (or
+The goal of real teaching is for the student to find the answer. And also the problem.
191
5.3 Summation versus Integration
1
2
3=n
Ax
1
2
3 = nAx
Fig. 5.6 Rectangles enclosing v = x2 have area (4n3+ in2 + AX)^ z
I
AX)^ = 3x3.
an integral). Algebra looks for f,'s whose differences produce v,. Here f, could be
called an antidiflerence (better to call it a sum).
The best start is a good guess. Copying directly from integrals, we might try
fn = fn3. To test if it is right, check whether f, -f n - I
produces on = n2:
We see n2, but also - n + f. The guess f n 3 needs correction terms. To cancel f in the
difference, I subtract f n from the sum. To put back n in the difference, I add
1 + 2 + .-.+ n = qn(n 1) to the sum. The new guess (which should be right) is
+
To check this answer, verify first that f l = 1. Also f2 = 5 and f3 = 14. To be certain,
verify that fn -f,-, = n2. For calculus the important term is in3:
n
j2 of the first n squares is
The sum
j= 1
1
1
1
n3 plus corrections - n2 and - n.
3
2
6
-
In practice f n 3 is an excellent estimate. The sum of the first 100 squares is approximately f(100)3, or a third of a million. If we need the exact answer, equation (7) is
available: the sum is 338,350. Many applications (example: the number of steps to
solve 100 linear equations) can settle for in3.
What is fascinating is the contrast with calculus. Calculus has no correction terms!
They get washed away in the limit of thin rectangles. When the sum is replaced by
the integral (the area), we get an absolutely clean answer:
The integral of v = x2 from x = 0 to x = n is exactly in3.
The area under the parabola, out to the point x = 100, is precisely a third of a million.
We have to explain why, with many rectangles.
The idea is to approach an infinite number of infinitely thin rectangles. A hundred
rectangles gave an area of 338,350. Now take a thousand rectangles. Their heights
are (&)2, (&)2, ... because the curve is v = x2. The base of every rectangle is
Ax = &, and we add heights times base:
area of rectangles =
(;J($) ($&)
+
+
* m e
+
(FJ(k).
Factor out (&)3. What you have left is l 2 + 22 + + 10002, which fits the sum of
squares formula. The exact area of the thousand rectangles is 333,833.5. I could try
to guess ten thousand rectangles but I won't.
Main point: The area is approaching 333,333.333. ... But the calculations are getting
worse. It is time for algebra-which means that we keep "Ax" and avoid numbers.
2
3
5
Integrals
The interval of length 100 is divided into n pieces of length Ax. (Thus n = 100/Ax.)
The jth rectangle meets the curve v = x2, so its height is AX)^. Its base is Ax, and
we add areas:
area = (AX)~(AX)
+ (2Ax)'(Ax)
+ ... + (nAx)'(Ax) =
n
j= 1
(jAx)'(Ax).
(8)
100
Factor out AX)^. leaving a sum of n squares. The area is (Ax)3 timesf., and n = -:
Ax
This equation shows what is happening. The leading term is a third of a million,
as predicted. The other terms are approaching zero! They contain Ax, and as the
rectangles get thinner they disappear. They only account for the small corners of
rectangles that lie above the curve. The vanishing of those corners will eventually be
proved for any continuous functions-the area from the correction terms goes to
zero-but here in equation (9) you see it explicitly.
The area under the curve came from the central idea of integration: 100/Ax rectanThe rectangular area is Z vj Ax.
gles of width Ax approach the limiting area = f
The exact area is j V(X)dx. In the limit Z becomes j and vj becomes v(x) and AX
becomes dx.
That completes the calculation for a parabola. It used the formula for a sum of
squares, which was special. But the underlying idea is much more general. The limit
of the sums agrees with the antiderivative: The antiderivative of v(x) = x2 isf(x) = i x 3 .
According to the Fundamental Theorem, the area under v(x) is f(x):
That Fundamental Theorem is not yet proved! I mean it is not proved by us. Whether
Leibniz or Newton managed to prove it, I am not quite sure. But it can be done.
Starting from sums of differences, the difficulty is that we have too many limits at
once. The sums of cjAx are approaching the integral. The differences Af/Ax approach
the derivative. A real proof has to separate those steps, and Section 5.7 will do it.
Proved or not, you are seeing the main point. What was true for the numbersf,
and cj is true in the limit for u(x) and.f(x). Now v(s) can vary continuously, but it is
still the slope of f'(s).The reverse of slope is area.
(1 + 2 + 3 + 412= 13 + 23 + 33 + 43
Proof without words by Roger Nelsen (Matlzenmtics
Finally we review the area under r; = x. The sum of 1 + 2 + + n is i n 2 + i n . This
gives the area of n = 4/Ax rectangles, going out to x = 4. The heights are jAx, the
bases are Ax, and we add areas:
193
5.3 Summation versus Integration
With A x = 1 the area is 1 + 2 + 3 + 4 = 10. With eight rectangles and Ax = f, the
area was 8 + 2Ax = 9. Sixteen rectangles of width i brought the correction 2Ax down
to f . The exact area is 8. The error is proportional to A x .
Important note There you see a question in applied mathematics. If there is an error,
what size is it? How does it behave as Ax + O? The A x term disappears in the limit,
and AX)^ disappears faster. But to get an error of
we need eight million
rectangles:
2A x = 2 4/8,000,000 = 10- 6.
That is horrifying! The numbers 10,9, 83, 8 i , ... seem to approach the area 8 in a
satisfactory way, but the convergence is much too slow. It takes twice as much work
to get one more binary digit in the answer-which is absolutely unacceptable. Somehow the A x term must be removed. If the correction is AX)^ instead of A x , then a
thousand rectangles will reach an accuracy of
The problem is that the rectangles are unbalanced. Their right sides touch the graph
of v, but their left sides are much too high. The best is to cross the graph in the middle
of the interval-this is the midpoint rule. Then the rectangle sits halfway across the
line v = x, and the error is zero. Section 5.8 comes back to this rule-and to Simpson's
rule that fits parabolas and removes the S AX)^ term and is built into many calculators.
Finally we try the quick way. The area under v = x is f = f x 2 , because dfldx is v.
The area out to x = 4 is 3(4)2= 8. Done.
Fig. 5.7 Endpoint rules: error
-
l/(work)
-
lln. Midpoint rule is better: error
-
l/(~ork)~.
Optional: pth powers Our sums are following a pattern. First, 1 + + n is f n2 plus
i n . The sum of squares is i n 3 plus correction terms. The sum of pth powers is
1 n P + l plus ~0wectionterms.
1~ + 2~ + ... + nP = ( 1 1)
p+l
The correction involves lower powers of n, and you know what is coming. Those
corrections disappear in calculus. The area under v = xP from 0 to n is
n/Ax
x p d x = lim
Ax+O j = 1
1
( ~ A x ) ~ ( A x-nP?
)=
~ +
l
Calculus doesn't care if the upper limit n is an integer, and it doesn't care if the power
p is an integer. We only need p + 1 > 0 to be sure nP+ is genuinely the leading term.
The antiderivative of v = xP is f = xP+' / ( p 1 ) .
We are close to interesting experiments. The correction terms disappear and the
sum approaches the integral. Here are actual numbers for p = 1, when the sum and
integral are easy: Sn= 1 + --.+ n and In= x dx = i n 2 . The difference is Dn= f n. The
thing to watch is the relative error En= Dn/In:
+
The number 20100 is f (200)(201). Please write down the next line n = 400, and please
jind a formula for En. You can guess En from the table, or you can derive it from
knowing Snand I,. The formula should show that En goes to zero. More important,
it should show how quick (or slow) that convergence will be.
One more number-a third of a million-was mentioned earlier. It came from
integrating x2 from 0 to 100, which compares to the sum Sloe of 100 squares:
These numbers suggest a new idea, to keep njixed and change p. The computer can
find sums without a formula! With its help we go to fourth powers and square roots:
lo0
$
67 1A629
3(100)~'~
4.7963
0.0072
In this and future tables we don't expect exact values. The last entries are rounded
off, and the goal is to see the pattern. The errors En,, are sure to obey a systematic
rule-they are proportional to l/n and to an unknown number C(p) that depends
on p. I hope you can push the experiments far enough to discover C(p). This is not
an exercise with an answer in the back of the book-it is mathematics.
6
Read-through questions
The Greek letter a indicates summation. In
uj the
dummy variable is b . The limits are c , so the first
. When uj = j this
term is d and the last term is
sum equals f . For n = 100 the leading term is g .
The correction term is h . The leading term equals the
integral of v = x from 0 to 100, which is written i . The
sum is the total i of 100 rectangles. The correction term
is the area between the k and the I .
The sum z:=, i2 is the same as 2;=,
m
and equals
n . The sum Z f = , vi is the same as 0
u i + , and equals
P . For& = Z;= vj the differencefn -f.- equals
4
.
Theformulafor 1 2 + 2 2 + . . . + n 2 i s f . = r .Toprove
it by mathematical induction, check f l = s and check
f.-S,- = t . The area under the parabola v = x2 from
x = 0 to x = 9 is
u . This is close to the area of
v
rectangles of base Ax. The correction terms approach zero
very w .
4
l/n and
3
j=O
( j 2-j) and
6
4 Evaluate
1 (-
2'.
i =0
i=O
n
1)'i and
1 (-
1 (2i - 3).
5 Write these sums in sigma notation and compute them:
6 Express these sums in sigma notation:
7 Convert these sums to sigma notation:
8 The binomial formula uses coefficients
100
9 With electronic help compute
j= 1
1 l/j
x
1
10
10 On a computer find
0
x
1000
and
1
n
1 112'.
1)'j.
j= 1
i= 1
i=2
n= 1
x
2' and
5
1 Compute the numbers
2 Compute
n
3 Evaluate the sum
10
(-l)'/i!
times
lo!
0
l/j.
195
5.4 Indefinite Integrals and Substitutions
x
n
11 Simplify
+ +
(ai bil2
i= 1
x
n
x
n
(ai - bi)2 to
a: and
i aibi# f aj i bk.
i= 1
x
n
13 "Telescope" the sums
(2'
- 2'-
j=1
k=1
29 The doubik sum
') and
J k= 1
v2 =
All but two terms cancel.
x
n
14 Simplify the sums
( 5 -5- 1) and
x (h+
j= 1
1
-5)-
30 he double sum
.
17 The antiderivative of d2fldx2 is dfldx. What is the sum
(f2 - 2fl+f0) + (f3 - 2f2 +fl) + "' + (f9 - 2f8 +f7)?
18 Induction: Verify that l2+ 22 + + n2 is f,=
n(n + 1)(2n+ 1)/6 by checking that fl is correct and
f,-f,-l = n2.
.-•
+ (2n - 1)= n2.
+ 23 + + n3 is f, = in2(n+
19 Prove by induction: 1 + 3 +
by checking f, and fn -f,-, . The text has a proof without words.
20 Verify that 1
+
21 Suppose f, has the form an bn2 + cn3. If you know
fl = 1, fi = 5, f3 = 14, turn those into three equations for
a, b, c. The solutions a = 4, b = 3, c = $ give what formula?
+ n8 = qn9+ correction.
23 Add n = 400 to the table for Sn= 1 + + n and find-the
22 Find q in the formula l8+
relative error En. Guess and prove a formula for En.
24 Add n = 50 to the table for Sn= l 2 + + n2 and compute
ESo.Find an approximate formula for En.
--•
25 Add p = 3 and p = 3 to the table for SloO,p=
1P + ---
+ 1 W . Guess an approximate formula for E1OO,p.
The
+
unless
.
+ v21 < lvll + lv21
]
(i +j) is vl =
x
(1 +j) plus
wi,j) is ( ~ 1 , 1 + ~ 1+,~21 , 3+
(j1
)
double
sum
i (i is
j=l
i = 1 wi,j)
. Compare.
(wl,l + ~2.1)+ h , 2 + ~2.2)+
31 Find the flaw in the proof that 2" = 1 for every
n = 0, 1,2, .... For n = 0 we have 2' = 1. If 2" = 1 for every
n e N, then 2N=2N-192N-1/2N-2=
1*1/1=1.
32 Write out all terms to see why the following are true:
+ +
33 The average of 6, 11, 4 is I7 = 3(6 11 4). Then
(6-@+(11-@+(4-fl=
. The average of
. Vl, ..., vn 1s v =
. Prove that Z (ui- 17)= 0.
34 The S ~ I W inequality
~ ~ Z is
($
aibiJ
<
($ ($
a:)
bf).
Compute both sides if al = 2, a2 = 3, bl = 1, b2 = 4. Then
compute both sides for any a,, a,, b,, b,. The proof in
Section 11.1 uses vectors.
35 Suppose n rectangles with base Ax touch the graph of v(x)
at the points x = Ax, 2Ax, ...,nAx. Express the total rectangular area in sigma notation.
36 If l/Ax rectangles with base Ax touch the graph of u(x)
at the left end of each interval (thus at x = 0, Ax, 2Ax, ...)
express the total area in sigma notation.
26 Guess C(p) in the formula E n ,z C(p)/n.
27 Show that 11 - 51 < 111 1-51. Always Ivl
x[
. None of this makes sense
(2 +j). Compute vl and v2 and the double sum.
12
j= 3
j= 1
+ of the (infinite)geometric
+ ... is the same as S minus
28 Let S be the sum 1 + x + x2
series. Then xS = x x2 + x3
. Therefore S =
if x = 2 because
+
i= 1
i= 1
37 The sum Ax
-
- 1)Ax)
1 f(jAx) -f((j
AX
'IAx
i= 1
In the limit this becomes 1;
5.4 Indefinite Integrals and Substitutions
equals
dx =
This section integrates the easy way, by looking for antiderivatives. We leave aside
sums of rectangular areas, and their limits as Ax -+ 0.Instead we search for an f (x)
with the required derivative u(x). In practice, this approach is more or less independent of the approach through sums-but it gives the same answer. And also, the
5 Integrals
search for an antiderivative may not succeed. We may not find f. In that case we go
back to rectangles, or on to something better in Section 5.8.
A computer is ready to integrate v, but not by discovering f . It integrates between
specified limits, to obtain a number (the definite integral). Here we hope to find a
function (the indefinite integral). That requires a symbolic integration code like
MACSYMA or Mathematica or MAPLE, or a reasonably nice v(x), or both. An
expression for f (x) can have tremendous advantages over a list of numbers.
Thus our goal is to find antiderivatives and use them. The techniques will be further
developed in Chapter 7-this section is short but good. First we write down what
we know. On each line, f (x) is an antiderivative of v(x) because df /dx = v(x).
Known pairs
Function v(x)
Antiderivative f (x)
Powers of x
xn
xn+'/(n + 1) + C
n = - 1 is not included, because n + 1 would be zero. v = x-' will lead us to f = In x.
cos x
sin x + C
sin x
-cos x + C
sec2x
tan x + C
sec x tan x
sec x + C
Trigonometricfunctions
Inverse functions
x +C
csc x cot x
- csc
I/,/-
sin-' x
1/(1 + x2)
tan-'
+C
x+C
You recognize that each integration formula came directly from a differentiation
formula. The integral of the cosine is the sine, because the derivative of the sine is
the cosine. For emphasis we list three derivatives above three integrals:
d
dx
- (constant) = 0
d
(x) = 1
dx
-
There are two ways to make this list longer. One is to find the derivative of a new
f (x). Then f goes in one column and v = df/dx goes in the other co1umn.T The other
possibility is to use rules for derivatives to find rules for integrals. That is the way to
extend the list, enormously and easily.
RULES FOR INTEGRALS
Among the rules for derivatives, three were of supreme importance. They were linearity, the product rule, and the chain rule. Everything flowed from those three. In the
t W e will soon meet ex, which goes in both columns. It is f ( x ) and also ~ ( x ) .
197
5.4 Indefinite Integrals and Substitutions
reverse direction (from v to f )this is still true. The three basic methods of differential
calculus also dominate integral calculus:
linearity of derivatives -,linearity of integrals
product rule for derivatives -+ integration by parts
chain rule for derivatives -+ integrals by substitution
The easiest is linearity, which comes first. Integration by parts will be left for
Section 7.1. This section starts on substitutions, reversing the chain rule to make an
integral simpler.
LINEARITY OF INTEGRALS
+
What is the integral of v(x) w(x)? Add the two separate integrals. The graph of
+ w has two regions below it, the area under v and the area from v to v + w.
Adding areas gives the sum rule. Suppose f and g are antiderivatives of v and w:
t.
sum rule:
constant rule:
linearity :
f
+g
is an antiderivative of
v+w
cf
is an antiderivative of
cv
af + bg is an antiderivative of av + bw
This is a case of overkill. The first two rules are special cases of the third, so logically
the last rule is enough. However it is so important to deal quickly with constantsjust "factor them outv-that the rule cv-cf is stated separately. The proofs come
from the linearity of derivatives: (af + bg)' equals af' + bg' which equals av + bw.
The rules can be restated with integral signs:
sum rule:
J [ ~ ( x+) w ( x ) ]d x = J
J CV(X) dx = c J
constant rule:
linearity:
V(X)
dx + J
V(X)
W ( X ) dx
dx
~[av(x)+bw(x)]dx=a~t.(x)dx+b~w(x)dx
Note about the constant in f ( x )+ C. All antiderivatives allow the addition of a constant. For a combination like av(x)+ bw(x), the antiderivative is af ( x )+ bg(x)+ C.
The constants for each part combine into a single constant. To give all possible antiderivatives of a function, just remember to write "+ C" after one of them. The real
problem is to find that one antiderivative.
EXAMPLE 1 The antiderivative of v = x2 + x -
is f
= x3/3
+ ( x - ')I(- 1) + C.
EXAMPLE 2 The antiderivative of 6 cos t + 7 sin t is 6 sin t - 7 cos t + C.
EXAMPLE 3 Rewrite
1
1 - sin x - 1 - sin x
as
1 + sin x
1 - sin2x
cos2x
= sec2x - sec x
tan x .
The antiderivative is tan x - sec x + C. That rewriting is done by a symbolic algebra
code (or by you). Differentiation is often simple, so most people check that df ldx = v(x).
Question How to integrate tan2 x?
Method Write it as sec2x - 1.
Answer
tan x
-
x
+C
5 Integrals
INTEGRALS BY SUBSTITUTION
We now present the most valuable technique in this section-substitution.
idea, you have to remember the chain rule:
To see the
f (g(x)) has derivative f '(g(x))(dg/dx)
sin x2 has derivative (cos x2)(2x)
(x3 + 1)'
has derivative 5(x3+ ll4(3x2)
If the function on the right is given, the function on the left is its antiderivative! There
are two points to emphasize right away:
1. Constants are no problem-they
can always be jixed. Divide by 2 or 15:
Notice the 2 from x2, the 5 from the fifth power, and the 3 from x3.
2. Choosing the insid?function g (or u) commits us to its derivative:
the integral of 2x cos x2 is sin x2 + C (g = x2, dgldx = 2x)
the integral of cos x2 is (failure)
(no dgldx)
the integral of x2 cos x2 is (failure)
(wrong dgldx)
To substitute g for x2, we need its derivative. The trick is to spot an inside function
whose derivative is present. We can fix constants like 2 or 15, but otherwise dgldx
has to be there. Very often the insidefunction g is written u. We use that letter to state
the substitution rule, when f is the integral of v :
EXAMPLE 4
EXAMPLE 5
EXAMPLE 6
1sin x cos x dx = &(sinx)' + C
1sin2x cos x dx = $(sin x ) +~ C
j cos x sin x dx = - f (cos x ) +~ C
u = sin x (compare Example 6)
u = sin x
u = cos x (compare Example 4)
The next example has u = x2 - 1 and duldx = 2x. The key step is choosing u:
EXAMPLE 8
xdx/,/n-
=
JFi+ C
j x J F T dx = $(x2- 1)3'2+ C
A ship of x (to x + 2) or a multiple of x (rescaling to 2x) is particularly easy:
EXAMPLES 9-40
5 (x + 2)) dx = $(x + 2)4 + C
j cos 2x dx = f sin 2x + C
You will soon be able to do those in your sleep. Officially the derivative of (x + 2)4
uses the chain rule. But the inside function u = x + 2 has duldx = 1. The "1" is there
automatically, and the graph shifts over-as in Figure 5.8b.
For Example 10 the inside function is u = 2x. Its derivative is duldx = 2. This
199
5.4 Indefinite Integrals and Substitutions
V(X -
xV (x 2)
area
lf
0
Fig. 5.8
(x2)
2
0
1
0
1
0
1
Substituting u = x + 1 and u = 2x and u = x 2. The last graph has half of du/dx = 2x.
required factor 2 is missing in J cos 2x dx, but we put it there by multiplying and
dividing by 2. Check the derivative of ½ sin 2x: the 2 from the chain rule cancels the
½. The rule for any nonzero constant is similar:
Sv(x + c) dx =f(x
+ c)
v(cx) dx= f (cx).
and
(2)
Squeezing the graph by c divides the area by c. Now 3x + 7 rescales and shifts:
EXAMPLE 14
f cos(3x+7) dx= ' sin(3x+7)+ C
(3x+7)3 + C
(3x+7)2 dx=
Remark on writing down the steps When the substitution is complicated, it is a good
idea to get du/dx where you need it. Here 3x2 + 1 needs 6x:
7x(3x 2 +
dx
Now integrate:
6
7 us
(3x2 + 1)46x dx
u4
dx
7 (3X 2 + 1)5
5 + C.
- - + C(3
65
6
(3)
5
Check the derivative at the end. The exponent 5 cancels 5 in the denominator, 6x from
the chain rule cancels 6, and 7x is what we started with.
Remark on differentials In place of (du/dx) dx, many people just write du:
S(3X
2
+ 1) 4 6x dx = u4 du = u5
+ C.
(4)
This really shows how substitution works. We switch from x to u, and we also switch
from dx to du. The most common mistake is to confuse dx with du. The factor du/dx
from the chain rule is absolutely needed, to reach du. The change of variables (dummy
variables anyway!) leaves an easy integral, and then u turns back into 3x 2 + 1. Here
are the four steps to substitute u for x:
1. Choose u(x) and compute du/dx
2. Locate v(u) times du/dx times dx, or v(u) times du
3. Integrate J v(u) du to find f(u) + C
4. Substitute u(x) back into this antiderivative f.
EXAMPLE 12
J(cos Vx) dxl 2/:
= f cos u du= sin u + C = sin
(put in u)
x+ C
(integrate) (put back x)
The choice of u must be right, to change everything from x to u. With ingenuity,
some remarkable integrals are possible. But most will remain impossible forever. The
functions cos x 2 and 1/ 4 - sin 2 x have no "elementary" antiderivative. Those integrals are well defined and they come up in applications--the latter gives the distance
5 Integrals
around an ellipse. That can be computed to tremendous accuracy, but not to perfect
accuracy.
The exercises concentrate on substitutions, which need and deserve practice. We
give a nonexample-1 (x2 + dx does not equal i ( x 2 + l)3-to emphasize the need
for duldx. Since 2x is missing, u = x2 1 does not work. But we can fix up n:
+
1
n
cos u + C = - - cos nx
n
+ C.
Read-through questions
dyldx = lly
26 dyldx = x/y
Finding integrals by substitution is the reverse of the a
rule. The derivative of (sin x ) is~ b . Therefore the antide~ x dx,
rivative of c is d . To compute (1 + sin x ) cos
so substitute
substitute u = e . Then duldx = f
h
=
I .
du = g . In terms of u the integral is
Returning to x gives the final answer.
d2y/dx2= 1
28 d y/dx5 = 1
d2y/dx2= - y
30 dy/dx =
d 2 ~ / d x=
2
32 (dyldx)' =
5
( 4 1v ( W ) dx =f (u(x))+ C
(b) J v2(x) dx = ff 3(x) C
(c) j v(x)(du/dx)dx =f ( ~ ( x ) ) C
(d) J v(x)(dv/dx)dx = 4f 2(x)+ C
+
5
(add
+ C)
2
1,/=
dx
+
+
(always C)
If
1cos3x sin x dx
9 1cos3 2x sin 2x dx
7
8
1cos x dx/sin3x
10 J cos3x sin 2x dx
1t , / g dt
14 1 t 3 & 7
dt
16 J (1 + x312)& dx
11
Jd
13
1t3 d t / J g
15
J (I + &)
17
J sec x tan x dx
18
19
1cos x tan x dx
20 J sin3x dx
t / J s
df /dx = v(x) then
v(x - I) dx =
and
df /dx = v(x) then
.
v(x2)xdx =
1v(2x - I) dx =
and
36 If
12
dx/&
j sec2x tan2x dx
In 21-32 find a function y(x) that solves the differential
equation.
21 dyldx = x2
+
True or false, when f is an antiderivative of v:
(a) J f(x)(dv/dx) dx =if2(x) C
(b) j v(v(x))(dvIdx)dx =f(V(X))
+C
(c) Integral is inverse to derivative so f (v(x))= x
(d) Integral is inverse to derivative so J (df /dx) dx =f (x)
Find the indefinite integrals in 1-20.
1J2$x. dx
&
True or false, when f is an antiderivative of v:
The best substitutions for 1tan (x + 3) sec2(x + 3)dx
and J ( ~ ~ + l ) ' ~ xare
d xu= I
and u = k . Then
du= I and m . The answers are n and 0 .
P .
2x dx/(l + x2)
The antiderivative of v dv/dx is
leads to J q , which we don't yet know. The integral
J dx/(l + x2) is known immediately as r .
1
fi
+ J;
23 dyldx = J1-Zx
22 dyldx = y 2 (try y = cxn)
24 dyldx = l / J n
+
38 j (x2 + 1)'dx is not &(x2 1)) but
39
J 2x dx/(x2+ 1) is J
.
du which will soon be In u.
40 Show that 12x3dx/(l
+x
~= j) (U~- 1) du/u3 =
41 The acceleration d2f /dt2 = 9.8 gives f (t) =
integration constants).
42 The solution to d 4 ~ / d x=
4 0 is
.
(two
(four constants).
43 If f(t) is an antiderivative of v(t), find antiderivatives of
(a) v(t
+ 3)
+
(b) v(t) 3 (c) 3v(t) (d) v(3t).
201
5.5 The Deflnlte Integral
The Definite lntearal
The integral of v(x) is an antiderivative f(x) plus a constant C. This section takes
two steps. First, we choose C. Second, we construct f (x). The object is to define the
integral-in the most frequent case when a suitable f (x) is not directly known.
The indefinite integral contains " + C." The constant is not settled because f (x) + C
has the same slope for every C. When we care only about the derivative, C makes
no difference. When the goal is a number-a definite integral-C can be assigned a
definite value at the starting point.
For mileage traveled, we subtract the reading at the start. This section does the
same for area. Distance is f(t) and area is f(x)-while the definite integral is
f (b) -f (a). Don't pay attention to t or x, pay attention to the great formula of integral
calculus:
lab Iab
~ ( tdt
) =
V(X)d~ =f (b) -f (a).
Viewpoint 1: When f is known, the equation gives the area from a to b.
Viewpoint 2: When f is not known, the equation defines f from the area.
For a typical v(x), we can't find f (x) by guessing or substitution. But still v(x) has an
"area" under its graph-and this yields the desired integral f (x).
Most of this section is theoretical, leading to the definition of the integral. You
may think we should have defined integrals before computing them-which is logically true. But the idea of area (and the use of rectangles) was already pretty clear in
our first examples. Now we go much further. Every continuous function v(x) has an
integral (also some discontinuous functions). Then the Fundamental Theorem completes the circle: The integral leads back to dfldx = u(x). The area up to x is the
antiderivative that we couldn't otherwise discover.
THE CONSTANT OF INTEGRATION
Our goal is to turn f (x) + C into a definite integral- the area between a and b. The
first requirement is to have area = zero at the start:
f (a) + C = starting area = 0 so C = -f (a).
(2)
For the area up to x (moving endpoint, indefinite integral), use t as the dummy variable:
the area from a to x is 1; v(t) dt =f (x) -f (a) (indefinite integral)
the a m a f r o a to b is
EXAMPLE I
v(x) dx =f (b) -f (a) (definite integral)
The area under the graph of 5(x + 1)4 from a to b has f (x) = (x + 1)':
The calculation has two separate steps-first find f (x), then substitute b and a. After
the first step, check that df /dx is v. The upper limit in the second step gives plus f (b),
the lower limit gives minus f(a). Notice the brackets (or the vertical bar):
f(x)]: =f(b)- f(a)
x31: = 8 - 1
Changing the example to f (x) = (x +
[cos
XI:'=cos 2t - 1.
- 1 gives an equally good antiderivative-
and now f (0)= 0. But f (b)-f (a)stays the same, because the - 1 disappears:
+
[ ( x + 1)' - 11: = ((b+ 1)' - 1) - ((a+ 1)' - 1)= (b + 1)' - (a 1)'.
EXAMPLE 2 When v = 2x sin x2 we recognize f = - cos x2. m e area from 0 to 3 is
The upper limit copies the minus sign. The lower limit gives - (- cos 0), which is
+ cos 0. That example shows the right form for solving exercises on dejkite integrals.
Example 2 jumped directly to f ( x )= - cos x2. But most problems involving the
chain rule go more slowly-by substitution. Set u = x2, with duldx = 2x:
IO3
2x sin x2 dx =
lo3
du
sin u - dx =
dx
sin u du.
We need new limits when u replaces x2. Those limits on u are a' and b2. (In this case
a' = O2 and b2 = 32 = 9.) Z f x goes from a to b, then u goesfrom ~ ( ato) u(b).
In this case u = x2 + 5. Therefore duldx = 2x (or du = 2x dx for differentials). We have
to account for the missing 2. The integral is Qu4. The limits on u = x2 + 5 are
u(0)= O2 + 5 and u(1)= 1' + 5. That is why the u-integral goes from 5 to 6. The
alternative is to find f ( x )= Q(x2+ 5)4 in one jump (and check it).
EXAMPLE 4
1: sin x2 dx = ?? (no elementaryfunction gives this integral).
If we try cos x2, the chain rule produces an extra 2x-no adjustment will work. Does
sin x2 still have an antiderivative? Yes! Every continuous v(x) has an f (x).Whether
f ( x ) has an algebraic formula or not, we can write it as J v(x)dx. To define that
integral, we now take the limit of rectangular areas.
INTEGRALS AS LIMITS OF "RIEMANN SUMS"
We have come to the definition of the integral. The chapter started with the integrals
of x and x2, from formulas for 1 + ..- + n and l 2 + ..- + n2. We will not go back to
those formulas. But for other functions, too irregular to find exact sums, the rectangular areas also approach a limit.
That limit is the integral. This definition is a major step in the theory of calculus.
It can be studied in detail, or understood in principle. The truth is that the definition
is not so painful-you virtually know it already.
Problem Integrate the continuousfunction v(x)over the interval [a, b].
Split [a, b] into n subintervals [a, x,], [x,, x 2 ] , ..., [xn- b].
Step 1
The "meshpoints" x,, x2, ... divide up the interval from a to b. The endpoints are
xo = a and x, = b. The length of subinterval k is Ax, = xk - xk- l . In that smaller
interval, the minimum of v(x) is mk.The maximum is M,.
203
5.5 The Definite Integral
Now construct rectangles. The "lower rectangle" over interval k has height mk. The
"upper rectangle" reaches to Mk. Since v is continuous, there are points Xmin and Xmax
where v = mk and v = Mk (extreme value theorem). The graph of v(x) is in between.
Important: The area under v(x) contains the area "s" of the lower rectangles:
fb v(x) dx > m
Ax1 + m2Ax 2 +
+ m,
nx,=
s.
(5)
The area under v(x) is contained in the area "S" of the upper rectangles:
fbv(x) dx
MAx + M2 Ax 2 +
+ MAxn= S.
(6)
The lower sum s and the upper sum S were computed earlier in special cases-when
v was x or x 2 and the spacings Ax were equal. Figure 5.9a shows why s < area < S.
A•v
v(x
(X)1
m
)V
(1
Mk
k
3,x
a
r.
"I·
Fig. 5.9
r.
"L·rl
h
-p.
r. X
,.
LI
nl,
Area of lower rectangles = s. Upper sum S includes top pieces. Riemann sum S* is in between.
Notice an important fact. When a new dividing point x' is added, the lower sum
increases. The minimum in one piece can be greater (see second figure) than the
original mk. Similarly the upper sum decreases. The maximum in one piece can be
below the overall maximum. As new points are added, s goes up and S comes down.
So the sums come closer together:
s < s'
<
IS'< S.
(7)
I have left space in between for the curved area-the integral of v(x).
Now add more and more meshpoints in such a way that Axmax -+ 0. The lower
sums increase and the upper sums decrease. They never pass each other. If v(x) is
continuous, those sums close in on a single number A. That number is the definite
integral-the area under the graph.
DEFINITION
The area A is the common limit of the lower and upper sums:
s - A and S -+ A as Axmax -+ 0.
(8)
This limit A exists for all continuous v(x), and also for some discontinuous functions.
When it exists, A is the "Riemann integral" of v(x) from a to b.
REMARKS ON THE INTEGRAL
As for derivatives, so for integrals: The definition involves a limit. Calculus is built
on limits, and we always add "if the limit exists." That is the delicate point. I hope
the next five remarks (increasingly technical) will help to distinguish functions that
are Riemann integrable from functions that are not.
Remark 1 The sums s and S may fail to approach the same limit. A standard
example has V(x) = 1 at all fractions x = p/q, and V(x) = 0 at all other points. Every
X
204
5 Integrals
interval contains rational points (fractions) and irrational points (nonrepeating decimals). Therefore mk = 0 and Mk = 1. The lower sum is always s = 0. The upper sum
is always S = b - a (the sum of 1's times Ax's). The gap in equation (7) stays open. This
function V(x) is not Riemann integrable. The area under its graph is not defined (at
least by Riemann-see Remark 5).
Remark 2 The step function U(x) is discontinuous but still integrable. On every
interval the minimum mk equals the maximum Mk-except on the interval containing
the jump. That jump interval has mk = 0 and Mk = 1. But when we multiply by Axk,
and require Axmax -+ 0, the difference between s and S goes to zero. The area under
a step function is clear-the rectangles fit exactly.
Remark 3 With patience another key step could be proved: If s -+ A and S -+ A for
one sequence of meshpoints, then this limit A is approached by every choice of meshpoints with Axmax
,
0. The integral is the lower bound of all upper sums S, and it is
the upper bound of all possible s-provided those bounds are equal. The gap must
close, to define the integral.
The same limit A is approached by "in-between rectangles." The height v(x*) can
be computed at any point x* in subinterval k. See Figures 5.9c and 5.10. Then the
total rectangular area is a "Riemann sum" between s and S:
S= v(x )Ax 1 + v(x*)Ax
2
+
...
+ v(x*)Ax.
(9)
We cannot tell whether the true area is above or below S*. Very often A is closer to
S* than to s or S. The midpoint rule takes x * in the middle of its interval (Figure 5.10),
and Section 5.8 will establish its extra accuracy. The extreme sums s and S are used
in the definition while S* is used in computation.
0-
/4""
V
(X
)/
right
Fig. 5.10
Remark 4
mid
./I
min
max
any x k
Various positions for x*' in the base. The rectangles have height v(x*).
Every continuous function is Riemann integrable. The proof is optional (in
my class), but it belongs here for reference. It starts with continuity at x*: "For any
e there is a 6 .... " When the rectangles sit between x* - 6 and x* + 6, the bounds Mk
and mk differ by less than 2e. Multiplying by the base Axk, the areas differ by less
than 2e(AXk). Combining all rectangles, the upper and lower sums differ by less than
2e(Ax 1 + Ax 2 + ... + Ax,)= 2e(b - a).
As e -+ 0 we conclude that S comes arbitrarily close to s. They squeeze in on a
single number A. The Riemann sums approach the Riemann integral, ifv is continuous.
Two problems are hidden by that reasoning. One is at the end, where S and s come
together. We have to know that the line of real numbers has no "holes," so there is
a number A to which these sequences converge. That is true.
Any increasingsequence, if it is bounded above, approaches a limit.
The decreasing sequence S, bounded below, converges to the same limit. So A exists.
The other problem is about continuity. We assumed without saying so that the
5.5 The Definite Integral
width 26 is the same around every point x*. We did not allow for the possibility that
6 might approach zero where v(x) is rapidly changing-in which case an infinite
number of rectangles could be needed. Our reasoning requires that
v(x) is unifomly continuous: 6 depends on E but not on the position of x*.
For each E there is a 6 that works at all points in the interval. A continuousfunction
on a closed interval is uniformly continuous. This fact (proof omitted) makes the
reasoning correct, and v(x) is integrable.
On an infinite interval, even v = x2 is not uniformly continuous. It changes across
~ - 6)2= 4x*6. As x* gets larger, 6 must get smallera subinterval by (x* + ~ 5-) (x*
to keep 4x*6 below E. No single 6 succeeds at all x*. But on a finite interval [O, b],
the choice 6 = ~ / 4 bworks everywhere-so v = x2 is uniformly continuous.
Remark 5 If those four remarks were fairly optional, this one is totally at your
discretion. Modern mathematics needs to integrate the zero-one function V(x) in the
first remark. Somehow V has more 0's than 1's. The fractions (where V(x) = 1) can
be put in a list, but the irrational numbers (where V(x) = 0) are "uncountable." The
integral ought to be zero, but Riemann's upper sums all involve M , = 1.
Lebesgue discovered a major improvement. He allowed infinitely many subintervals
(smaller and smaller). Then all fractions can be covered with intervals of total width
E. (Amazing, when the fractions are packed so densely.) The idea is to cover 1/q, 2/q,
...,q/q by narrow intervals of total width ~ 1 2Combining
~.
all q = 1,2, 3, ...,the total
width to cover all fractions is no more than E(&+ $ + $ + --.)= E. Since V(x) = 0
everywhere else, the upper sum S is only E. And since E was arbitrary, the "Lebesgue
integral" is zero as desired.
That completes a fair amount of theory, possibly more than you want or needbut it is satisfying to get things straight. The definition of the integral is still being
studied by experts (and so is the derivative, again to allow more functions). By
contrast, the properties of the integral are used by everybody. Therefore the next
section turns from definition to properties, collecting the rules that are needed in
applications. They are very straightforward.
5.5 EXERCISES
Read-through questions
In J: v(t) dt =f (x) + C, the constant C equals a . Then
at x = a the integral is b . At x = b the integral becomes
c . The notation f ($1: means d . Thus cos x]: equals
e . Also [cos x + 3]",quals
t , which shows why
the antiderivative includes an arbitrary Q . Substituting
u = 2x - 1 changes J:
dx into h (with limits
on u).
Jn
The integral J,b U(X)dx can be defined for any I function v(x), even if we can't find a simple i . First the meshpoints xl, x2, ... divide [a, b] into subintervals of length
Axk= k . The upper rectangle with base Ax, has height
Mk= 1 . The upper sum S is equal to m . The lower
sum s is n . The o is between s and S. As more
meshpoints are added, S P and s q . If S and s
approach the same r ,that defines the integral. The intermediate sums S*, named after s , use rectangles of height
v(x,*). Here X$ is any point between t , and S* = u
approaches the area.
If u(x) = dfldx, what constants C make 1-10 true?
1
Jb,
V(X)dx =f (b) + C
2 j; v(x) dx =f (4) + C
3
1: v(t) dt = -f (x) + C
4
J:,
5
v(sin x) cos x dx =f (sin b) + C
v(t) dt =f (t) + C (careful)
6 dfldx = v(x)
7
+C
1; (x2-l)j2x
dx=j:,
u3du.
8
I:' v(t) dt =f (x2)+ C
26 Find the Riemann sum S* for V(x) in Remark 1, when
1: v(- X)dx = C (change -x to t; also dx and limits)
10 1
; v(x) dx = C v(2t) dt.
9
Ax = l/n and each xf is the midpoint. This S* is well-behaved
but still V(x) is not Riemann integrable.
27 W(x) equals S at x = 3,4,4, ..., and elsewhere W(x) = 0.
For Ax = .O1 find the upper sum S. Is W(x) integrable?
Choose u(x) in 11-18 and change limits. Compute the integral
in 11-16.
1; (x2+ l)lOxdx
13 El4tan x sec2x dx
11
15
17
sec2'xtan x dx
1: dx/x
(take u = l/x)
1": sin8x cos x dx
14 1; x2"+' dx (take u = x2)
16 1
; xdx/Jm'
18 1
; x3(1- x ) dx
~ (u = 1 - x)
12
With Ax = 3 in 19-22, find the maximum Mk and minimum
mk and upper and lower sums S and s.
19
21
1; (x' + 114dx
x3 dx
20
28 Suppose M(x) is a multistep function with jumps of 3, f ,
4, ... at the points x = +,&,
4, .... Draw a rough graph with
M(0) = 0 and M(1) = 1. With Ax = 5 find S and s.
29 For M(x) in Problem 28 find the difference S - s (which
approaches zero as Ax -* 0). What is the area under the
graph?
30 If dfldx = - V(X)and f(I) = 0, explain f (x) =
+ v(x) and f (0) = 3, find f (x).
(b) If df /dx = + v(x) and f (3) = 0, find f (x).
31 (a) If df /dx =
32 In your own words define the integral of v(x) from a to b.
33 True or false, with reason or example.
sin 2nx dx
22
x dx.
23 Repeat 19 and 20 with Ax = 4 and compare with the cor-
rect answer.
rectangle. Find Ax so that S < 4.001.
25 If v(x) is increasing for a ,< x ,< b, the difference S - s is the
1
5.6
(a) Every continuous v(x) has an antiderivative f (x).
(b) If v(x) is not continuous, S and s approach different
limits.
(c) If S and s approach A as Ax + 0, then all Riemann
sums S* in equation (9) also approach A.
(d) If vl(x) v2(x)= u3(x), their upper sums satisfy
S1 +S2 =S3.
(e) If vl(x) v2(x)= u3(x), their Riemann sums at the
midpoints xf satisfy Sf + S t = ST.
(f) The midpoint sum is the average of S and s.
(g) One xf in Figure 5.10 gives the exact area
+
+
24 The difference S - s in 21 is the area 23Ax of the far right
area of the
rectangle minus the area of the
rectangle. Those areas approach zero. So every increasing
function on [a, b] is Riemann integrable.
1: v(t) dt.
'
Properties of the Integral and Average Value
m The previous section reached the definition of 1: v(x) dx. But the subject cannot stop
there. The integral was defined in order to be used. Its properties are important, and
its applications are even more important. The definition was chosen so that the
integral has properties that make the applications possible.
One direct application is to the average value of v(x). The average of n numbers is
clear, and the integral extends that idea-it produces the average of a whole continuum of numbers v(x). This develops from the last rule in the following list (Property
7). We now collect together seven basic properties of defirrite integrals.
The addition rule for [v(x) + w(x)] dx will not be repeated-even though this
property of linearity is the most fundamental. We start instead with a different kind
of addition. There is only one function v(x), but now there are two intervals.
The integralfrom a to b is added to its neighborfrom b to c. Their sum is the integral
from a to c. That is the first (not surprising) property in the list.
207
5.6 Properties of the Integral and Average Value
Property 1 Areas over neighboring intervals add to the area over the combined
interval:
v(x) dx + I' v(x) dx = J v(x) dx.
(1)
J
This sum of areas is graphically obvious (Figure 5.1 la). It also comes from the formal
definition of the integral. Rectangular areas obey (1)-with a meshpoint at x = b to
make sure. When Axmax approaches zero, their limits also obey (1). All the normal
rules for rectangularareas are obeyed in the limit by integrals.
Property 1 is worth pursuing. It indicates how to define the integral when a = b.
The integral "from b to b" is the area over a point, which we expect to be zero. It is.
fb v(x) dx = 0.
Property 2
That comes from Property 1 when c = b. Equation (1) has two identical integrals, so
the one from b to b must be zero. Next we see what happens if c = a-which makes
the second integral go from b to a.
What happens when an integralgoes backward? The "lower limit" is now the larger
number b. The "upper limit" a is smaller. Going backward reverses the sign:
fa v(x)
Property 3
dx = -
f~ v(x) dx =f(a)-f(b).
Proof When c = a the right side of (1) is zero. Then the integrals on the left side
must cancel, which is Property 3. In goingfrom b to a the steps Ax are negative. That
justifies a minus sign on the rectangular areas, and a minus sign on the integral
(Figure 5.1 1b). Conclusion: Property 1 holds for any ordering of a, b, c.
EXAMPLES
t2 dt = - -
dt = -1
_=0
Property 4 For odd functions Ja, v(x) dx = O0."Odd" means that v(- x) = - v(x).
For even functions •-a v(x) dx = 2 fo v(x) dx. "Even" means that v(- x) = + v(x).
The functions x, x3 , x 5, ... are odd. If x changes sign, these powers change sign. The
functions sin x and tan x are also odd, together with their inverses. This is an important family of functions, and the integralof an odd function from - a to a equals zero.
Areas cancel:
j•a
6x d= x]',
a6 -(- a)6 = 0.
If v(x) is odd then f(x) is even! All powers 1, x 2, x4 ,... are even functions. Curious
fact: Odd function times even function is odd, but odd number times even number is
even.
For even functions, areas add: J"a cos x dx = sin a - sin(- a) = 2 sin a.
v(-x) = - v(x)
a
-
o _
c
a
-1
o
x
-x
x
I
Fig. 5.11 Properties 1-4: Add areas, change sign to go backward, odd cancels, even adds.
Integrals
5
The next properties involve inequalities. If v(x) is positive, the area under its graph
is positive (not surprising). Now we have a proof: The lower sums s are positive and
they increase toward the area integral. So the integral is positive:
Property 5 If v(x) > 0 for a < x < b then J v(x) dx > 0.
A positive velocity means a positive distance. A positive v lies above a positive area.
A more general statement is true. Suppose v(x) stays between a lower function 1(x)
and an upper function u(x). Then the rectangles for v stay between the rectangles for 1
and u. In the limit, the area under v (Figure 5.12) is between the areas under I and u:
Property 6
If 1(x) < v(x) < u(x) for a < x < b then
1(x) dx
II
~a v(x) dx
EXAMPLE 1
cos t<1
=~
cosCt dt I
EXAMPLE 2
1 sec 2 t
=•
1 dt <• sec 2 tdtdt
EXAMPLE 3 Integrating 1
1
2
1 dt
~a u(x) dx.
=
sin x
(2)
x
x <tanx
1 leads to tan- x < x.
All those examples are for x > 0. You may remember that Section 2.4 used geometry
to prove sin h < h < tan h. Examples 1-2 seem to give new and shorter proofs. But I
think the reasoning is doubtful. The inequalities were needed to compute the derivatives (therefore the integrals) in the first place.
Vave
Fig. 5.12 Properties 5-7: v above zero, v between 1and u, average value (+ balances -).
Property 7 (Mean Value Theorem for Integrals) If v(x) is continuous, there is a
point c between a and b where v(c) equals the average value of v(x):
(cv(c)b-a I a v(x) dx = "average value of v(x)."
(3)
This is the same as the ordinary Mean Value Theorem (for the derivative of f(x)):
f'(c) -
f(b) -f(a)
b-a (a)-
average slope of f."
(4)
With f' = v, (3) and (4) are the same equation. But honesty makes me admit to a flaw
in the logic. We need the Fundamental Theorem of Calculus to guarantee that
f(x) = f v(t) dt really gives f'= v.
A direct proof of (3)places one rectangle across the interval trom a to b. Now raise
the top of that rectangle, starting at Vmin (the bottom of the curve) and moving up to
vmax (the top of the curve). At some height the area will be just right-equal to the
area under the curve. Then the rectangular area, which is (b - a) times v(c), equals
the curved area Jf v(x) dx. This is equation (3).
5.6 Properties of the Integral and Average Value
/
u(x)=x
u(x) = sin2x
u(x>= x2
Fig. 5.13 Mean Value Theorem for integrals: area/(b - a) = average height
= v(c)
at some c.
That direct proof uses the intermediate value theorem: A continuous function v(x)
takes on every height between v,, and v,,,. At some point (at two points in
Figure 5.12~)the function v(x) equals its own average value.
Figure 5.13 shows equal areas above and below the average height v(c) = vaVe.
EXAMPLE 4 The average value of an odd function is zero (between
-
1 and 1):
For once we know c. It is the center point x = 0, where v(c) = vav, = 0.
EXAMPLE 5 The average value of x2 is f (between 1 and - 1):
(note
,,-7
-
Where does this function x2 equal its average value f? That happens when c2 = f , so
c can be either of the points I/& and - 1/J? in Figure 5.13b. Those are the Gauss
points, which are terrific for numerical integration as Section 5.8 will show.
EXAMPLE 6 The average value of sin2 x over a period (zero to n) is i :
(note
- ;7
-
The point c is n/4 or 344, where sin2 c = $. The graph of sin2x oscillates around its
average value f . See the figure or the formula:
sin2 x = f - f cos 2x.
(5)
The steady term is f , the oscillation is - 4 cos 2x. The integral is f (x) = i x - sin 2x,
which is the same as f x - i sin x cos x. This integral of sin2 x will be seen again. Please
verify that df /dx = sin2x.
THE AVERAGE VALUE AND EXPECTED VALUE
The "average value" from a to b is the integral divided by the length b - a. This
was computed for x and x2 and sin2 x, but not explained. It is a major application
of the integral, and it is guided by the ordinary average of n numbers:
Vave
=-
V(X)dx
comes from
1
uave = - (vl + v2 +
n
... + v,).
Integration is parallel to summation! Sums approach integrals. Discrete averages
5 Integrals
approach continuous averages. The average of 4, %,
3 is 3. The average of f ,$,3, 4,
3 is 3. The average of n numbers from l/n to n/n is
The middle term gives the average, when n is odd. Or we can do the addition. As
n -,oo the sum approaches an integral (do you see the rectangles?). The ordinary
average of numbers becomes the continuous average of v(x) = x:
n + l + -1
2n
2
and
Iol
x dx =
(note
b-o -1
)
In ordinary language: "The average value of the numbers between 0 and 1 is 4." Since
a whole continuum of numbers lies between 0 and 1, that statement is meaningless
until we have integration.
The average value of the squares of those numbers is (x2),,, = x2dx/(b - a) = 4.
Ifyou pick a number randomly between 0 and 1, its expected value is 5 and its expected
square is 3.
To me that sentence is a puzzle. First, we don't expect the number to be exactly
&so we need to define "expected value." Second, if the expected value is 9, why is
the expected square equal to 3 instead of i?The ideas come from probability theory,
and calculus is leading us to continuous probability. We introduce it briefly here, and
come back to it in Chapter 8.
PREDlClABLE AVERAGES FROM RANDOM EVENTS
Suppose you throw a pair of dice. The outcome is not predictable. Otherwise why
throw them? But the average over more and more throws is totally predictable. We
don't know what will happen, but we know its probability.
For dice, we are adding two numbers between 1 and 6. The outcome is between 2
and 12. The probability of 2 is the chance of two ones: (1/6)(1/6)= 1/36. Beside each
outcome we can write its probability:
To repeat, one roll is unpredictable. Only the probabilities are known, and they add
to 1. (Those fractions add to 36/36; all possibilities are covered.) The total from a
million rolls is even more unpredictable-it can be anywhere between 2,000,000 and
12,000,000. Nevertheless the average of those million outcomes is almost completely
predictable. This expected value is found by adding the products in that line above:
Expected value: multiply (outcome)times (probability of outcome) and add:
If you throw the dice 1000 times, and the average is not between 6.9 and 7.1, you get
an A. Use the random number generator on a computer and round off to integers.
Now comes continuous probability. Suppose all numbers between 2 and 12 are
equally probable. This means all numbers-not just integers. What is the probability
of hitting the particular number x = n? It is zero! By any reasonable measure, n has
5.6 Properties of the Integral and Average Value
no chance to occur. In the continuous case, every x has probability zero. But an
interval of x's has a nonzero probability:
the probability of an outcome between 2 and 3 is 1/10
the probability of an outcome between x and x + Ax is Ax110
To find the average, add up each outcome times the probability of that outcome.
First divide 2 to 12 into intervals of length Ax = 1 and probability p = 1/10. If we
round off x, the average is 63:
Here all outcomes are integers (as with dice). It is more accurate to use 20 intervals
of length 112 and probability 1/20. The average is 6$, and you see what is coming.
These are rectangular areas (Riemann sums). As Ax -+ 0 they approach an integral.
The probability of an outcome between x and x + dx is p(x) dx, and this problem has
p(x) = 1/10. The average outcome in the continuous case is not a sum but an integral:
expected value E(x) =
dx x2
xp(x) dx = S212 x 10= 20]2
l2
= 7.
That is a big jump. From the point of view of integration, it is a limit of sums. From
the point of view of probability, the chance of each outcome is zero but the probability
density at x is p(x) = 1/10. The integral of p(x) is 1, because some outcome must
happen. The integral of xp(x) is x,,, = 7, the expected value. Each choice of x is
random, but the average is predictable.
This completes a first step in probability theory. The second step comes after more
calculus. Decaying probabilities use e-" and e-"'-then
the chance of a large x is
very small. Here we end with the expected values of xn and I/& and l/x, for a
random choice between 0 and 1 (so p(x) = 1):
A CONFUSION ABOUT "EXPECTED" CLASS SIZE
A college can advertise an average class size of 29, while most students are in large
classes most of the time. I will show quickly how that happens.
Suppose there are 95 classes of 20 students and 5 classes of 200 students. The total
enrollment in 100 classes is 1900 + 1000 = 2900. A random professor has expected
class size 29. But a random student sees it differently. The probability is 1900/2900
of being in a small class and 1000/2900 of being in a large class. Adding class size
times probability gives the expected class size for the student:
(20) (E)
+ (200) (IWO)
2900
2900
= 82
students in the class.
Similarly, the average waiting time at a restaurant seems like 40 minutes (to the
customer). To the hostess, who averages over the whole day, it is 10 minutes. If you
came at a random time it would be 10, but if you are a random customer it is 40.
Traffic problems could be eliminated by raising the average number of people per
car to 2.5, or even 2. But that is virtually impossible. Part of the problem is the
5 Integrals
difference between (a) the percentage of cars with one person and (b) the percentage
of people alone in a car. Percentage (b) is smaller. In practice, most people would be
in crowded cars. See Problems 37-38.
17 What number 8 gives !j (v(x)- 8) dx = O?
Read-through questions
1;
The integrals v(x) dx and
v(x) dx add to a . The
integral
v(x) dx equals b . The reason is c . If
V(X)< x then v(x) dx < d . The average value of v(x) on
the interval 1 < x < 9 is defined by
. It is equal to u(c)
at a point x = c which is f . The rectangle across this
interval with height v(c) has the same area as g . The
average value of u(x) = x + 1 on the interval 1 < x < 9 is
h
If x is chosen from 1, 3, 5, 7 with equal probabilities $, its
expected value (average) is 1 . The expected value of x2
is 1 . If x is chosen from 1, 2, ..., 8 with probabilities i,
its expected value is k . If x is chosen from 1 < x < 9, the
chance of hitting an integer is I . The chance of falling
between x and x + dx is p(x) dx = m . The expected value
E(x) is the integral n . It equals 0 .
In 1-6 find the average value of v(x) between a and b, and find
all points c where vave = v(c).
18 If f (2) = 6 and f (6) = 2 then the average of df /dx from
.
x=2tox=6is
19 (a) The averages of cos x and lcos xl from 0 to n are
...,v,
is
than
20 (a) Which property of integrals proves
ji v(x) dx <
(b) The average of the numbers v,,
the average of Ivll, ..., lu,l.
j: I.(x,I dx?
(b) Which property proves
-1: v(x) dx < j: Iv(x)l dx?
Together these are Property 8: 11;v(x) d x l 6 Iv(x)l dx.
21 What function has vave (from 0 to x) equal to $ v(x) at all
x? What functions have vave = v(x) at all x?
22 (a) If v(x) is increasing, explain from Property 6 why
j",(t)
dt < xv(x) for x > 0.
(b) Take derivatives of both sides for a second proof.
23 The average of v(x) = 1/(1+ x 2 ) on the interval [0, b]
as b -+ co. The average of V(x) =
approaches
x2/(1+ x2) approaches
.
24 If the positive numbers v, approach zero as n -+ co prove
that their average (vl + - - - + vJn also approaches zero.
25 Find the average distance from x = a to points in the
interval 0 < x < 2. Is the formula different if a < 2?
26 (Computer experiment) Choose random numbers x
Are 9-16 true or false? Give a reason or an example.
9 The minimum of
10 The value of
S", v(t) dt is at x = 4.
v(t) dt does not depend on x.
11 The average value from x = 0 to x = 3 equals
$(vaVeon 0 < x < 1) + 3(vav,on 1 < x < 3).
12 The ratio (f (b) -f (a))/(b- a) is the average value of f (x)
ona<x<b.
13 On the symmetric interval -1
< x < 1, v(x) - vave is an
odd function.
14 If l(x) < v(x) < u(x) then dlldx < dvldx < duldx.
15 The average of v(x) from 0 to 2 plus the average from 2
to 4 equals the average from 0 to 4.
16 (a) Antiderivatives of even functions are odd functions.
(b) Squares of odd functions are odd functions.
between 0 and 1 until the average value of x2 is between .333
and .334. How many values of x2 are above and below? If
possible repeat ten times.
27 A point P is chosen randomly along a semicircle (see
figure: equal probability for equal arcs). What is the
average distance y from the x axis? The radius is 1.
28 A point Q is chosen randomly between -1 and 1.
(a) What is the average distance Y up to the semicircle?
(b) Why is this different from Problem 27?
Buffon needle
213
5.7 The Fundamental Theorem and Its Applications
29 (A classic way to compute n;) A 2" needle is tossed onto
a floor with boards 2" wide. Find the probability of falling
across a crack. (This happens when cos 8> y = distance from
midpoint of needle to nearest crack. In the rectangle
0 6 8 < 7r/2,O < y 6 1, shade the part where cos 8 > y and find
the fraction of area that is shaded.)
30 If Buffon's needle has length 2x instead of 2, find the
probability P(x) of falling across the same cracks.
31 If you roll three dice at once, what are the probabilities of
each outcome between 3 and 18? What is the expected value?
32 If you choose a random point in the square 0 6 x < 1,
0 < y 6 1, what is the chance that its coordinates have yZ < x?
33 The voltage V(t) = 220 cos 2n;t/60 has frequency 60 hertz
and amplitude 220 volts. Find Kvefrom 0 to t.
34 (a) Show that veve,(x)= $(v(x) + u(-x)) is always even.
= $(v(x)- v(-x)) is always odd.
(b) Show that vOdd(x)
+
35 By Problem 34 or otherwise, write (x
as an even function plus an odd function.
and l/(x + 1)
36 Prove from the definition of dfldx that it is an odd function if f (x) is even.
37 Suppose four classes have 6,8,10, and 40 students, averag. The chance of being in the first class is
ing
. The expected class size (for the student) is
38 With groups of sizes xl , ...,x, adding to G, the average
. The chance of an individual belonging to
size is
group 1 is
. The expected size of his or her group is
E(x) = x, (xl /G) + -.- + x,(x,/G). *Prove Z: X?/G 2 G/n.
True or false, 15 seconds each:
(a) If f (x) < g(x) then df ldx 6 dgldx.
(b) If df /dx 6 dgldx then f (x) < g(x).
(c) xv(x) is odd if v(x) is even.
(d) If v,, d waveon all intervals then u(x) 6 w(x) at all
points.
If v(x) =
Thus
x2 for x < 3
2x for x < 3
then f(x) =
-2x for x > 3
-x2 for x > 3 '
v(x) dx =f (4) -f (0) = - 16. Correct the mistake.
41 If v(x) = Ix - 2) find f (x). Compute
-
u(x) dx.
42 Why are there equal areas above and below vave?
5.7 The Fundamental Theorem and Its Applications
When the endpoints are fixed at a and b, we have a definite integral. When the upper
limit is a variable point x, we have an indefinite integral. More generally: When the
endpoints depend in any way on x, the integral is a function of x. Therefore we can
find its derivative. This requires the Fundamental Theorem of Calculus.
The essence of the Theorem is: Derivative of integral of v equals v. We also compute
the derivative when the integral goes from a(x) to b(x)-both limits variable.
Part 2 of the Fundamental Theorem reverses the order: Integral ofderivative o f f
equals f C . That will follow quickly from Part 1, with help from the Mean Value
Theorem. It is Part 2 that we use most, since integrals are harder than derivatives.
After the proofs we go to new applications, beyond the standard problem of area
under a curve. Integrals can add up rings and triangles and shells-not just rectangles.
The answer can be a volume or a probability-not just an area.
+
THE FUNDAMENTAL THEOREM, PART 1
Start with a continuous function v . Integrate it from a fixed point a to a variable
point x. For each x, this integral f(x) is a number. We do not require or expect a
formula for f (x)-it is the area out to the point x. It is a function of x! The Fundamental Theorem says that this area function has a derivative (another limiting process).
The derivative df ldx equals the original v(x).
5 Integrals
The dummy variable is written as t, so we can concentrate on the limits. The val
of the integral depends on the limits a and x, not on t.
To find df ldx, start with Af =f (x + Ax) -f (x) = diflerence of areas:
I."+Ax~ ( tdt) - 1; v(t) dt =
v(t) dt.
(1)
Officially, this is Property 1. The area out to x + Ax minus the area out to x equals
~f=
the small part from x to x + Ax. Now divide by Ax:
1 x+Ax
Af--v(t) dt = average value = v(c).
Ax Ax
I
This is Property 7, the Mean Value Theorem for integrals. The average value on this
short interval equals v(c). This point c is somewhere between x and x + Ax (exact
position not known), and we let Ax approach zero. That squeezes c toward x, so v(c)
approaches u(x)-remember that v is continuous. The limit of equation (2) is the
Fundamental Theorem:
Af + d f
Ax
dx
and v(c) + u(x)
SO
df
dx
- = v(x).
If Ax is negative the reasoning still holds. Why assume that v(x) is continuous?
Because if v is a step function, then f (x) has a corner where dfldx is not v(x).
We could skip the Mean Value Theorem and simply bound v above and below:
for t between x and x + Ax:
integrate over that interval:
umin
6 ~ ( t )G Vmax
vminAxQ Af
G vmaxAx
tp
(4)
As Ax -,0, umin and vmax approach v(x). In the limit dfldx again equals v(x).
j.(.\-+ A.v)
f(.d
x
X+AK
Af *= u(.u)A.r
x
.\-+AX
Fig. 5.14 Fundamental Theorem, Part 1: (thin area Af)/(base length Ax) -+ height u(x).
Graphical meaning The f-graph gives the area under the v-graph. The thin strip in
Figure 5.14, has area Af. That area is approximately v(x) times Ax. Dividing by its
base, AflAx is close to the height v(x). When Ax -* 0 and the strip becomes infinitely
thin, the expression "close to" converges to "equals." Then df ldx is the height at v(x).
DERIVATIVES WITH VARIABLE ENDPOINTS When the upper limit is x, the derivative is v(x). Suppose the lower limit is x. The
integral goes from x to 6,instead of a to x. When x moves, the lower limit moves.
5.7 The Fundamental Theorem and Its Applications
The change in area is on the left side of Figure 5.15. As x goesforward, area is removed.
So there is a minus sign in the derivative of area:
The derivative of g(x) =
dg = - v(x).
v(t) dt is dx
The quickest proof is to reverse b and x, which reverses the sign (Property 3):
g(x) = -
1'
v(t) dt so by part I dg = - v(x).
dx
Fig. 5.15 Area from x to b has dgldx = - u(x). Area v(b)db is added, area v(a)da is lost
The general case is messier but not much harder (it is quite useful). Suppose both
limits are changing. The upper limit b(x) is not necessarily x, but it depends on x.
The lower limit a(x) can also depend on x (Figure 5.15b). The area A between those
limits changes as x changes, and we want dAldx:
v(t) dt then
dA
dx
-=
db
da
v(b(x)) - - v(a(x)) -.
dx
dx
The figure shows two thin strips, one added to the area and one subtracted.
First check the two cases we know. When a = 0 and b = x, we have daldx = 0 and
dbldx = 1. The derivative according to (6) is v(x) times 1-the Fundamental Theorem.
The other case has a = x and b = constant. Then the lower limit in (6) produces - v(x).
When the integral goes from a = 2x to b = x3, its derivative is new:
EXAMPLE 1
A
=
5;:
cos t dt = sin x3 - sin 2x
dAjdx = (cos x3)(3x2)- (cos 2x)(2).
That fits with (6), because dbldx is 3x2 and daldx is 2 (with minus sign). It also looks
like the chain rule-which it is! To prove (6) we use the letters v and f:
A=
~ ( t dt
) =j(h(x)) -f (a(x))
(by Part 2 below)
(by the chain rule)
Since f ' = v, equation (6) is proved. In the next example the area turns out to be
constant, although it seems to depend on x. Note that v(t) = l/t so v(3x) = 1/3x.
EXAMPLE2 A=[:
- dt
t
has dA =
dx
( ) (&)
(3) -
(2) = 0.
5
Question A =
I.
u(t) dt has
Integrals
dA
- = u(x)
dx
+ v(-
x). Why does v(- x) have a plus sign?
THE FUNDAMENTAL THEOREM, PART 2 We have used a hundred times the Theorem that is now to be proved. It is the key
to integration. "The integral of dfldx is f (x) + C." The application starts with v(x).
We search for an f (x) with this derivative. If dfldx = v(x), the Theorem says that
We can't rely on knowing formulas for v and f-only the definitions of and dldx.
The proof rests on one extremely special case: dfldx is the zero function. We easily
find f (x) = constant. The problem is to prove that there are no other possibilities: f '
must be constant. When the slope is zero, the graph must be flat. Everybody knows
this is true, but intuition is not the same as proof.
Assume that df ldx = 0 in an interval. Iff (x) is not constant, there are points where
f (a) # f (b). By the Mean Value Theorem, there is a point c where
f '(c) = (b) -f
b-a
(this is not zero because f (a) # f (b)).
But f '(c) # 0 directly contradicts df ldx = 0. Therefore f (x) must be constant.
Note the crucial role of the Mean Value Theorem. A local hypothesis (dfldx = 0
at each point) yields a global conclusion (f = constant in the whole interval). The
derivative narrows the field of view, the integral widens it. The Mean Value Theorem
connects instantaneous to average, local to global, points to intervals. This special
case (the zero function) applies when A(x) and f(x) have the same derivative:
IfdAldx
= dfldx
on an interval, then A(x) =f(x) + C.
(7)
Reason: The derivative of A(x) -f (x) is zero. So A(x) -f (x) must be constant.
Now comes the big theorem. It assumes that v(x) is continuous, and integrates
using f (x):
5D (Fu~tdamentalTheorem, Part 2) If u(x) =
dx
u(x) dx =f (b) -f (a).
The antiderivative is f (x). But Part 1gave another antiderivative for the same
v(x). It was the integral-constructed from rectangles and now called A(x):
Proof
v(t)dt
dA
alsohas ---=v(x).
dx
Since A' = v and f ' = v, the special case in equation (7) states that A(x) =f (x) + C.
That is the essential point: The integral from rectangles equals f (x) + C.
At the lower limit the area integral is A = 0. So f (a) + C = 0. At the upper limit
j'(b) + C = A(b). Subtract to find A(b), the definite integral:
Calculus is beautiful-its
Fundamental Theorem is also its most useful theorem.
217
5.7 The Fundamental Theorem and Its Applications
Another proof of Part 2 starts with f' = v and looks at subintervals:
f(xi) - f(a) = v(x*)(xi - a)
(by the Mean Value Theorem)
f(x 2) -f(x 1)= V(X2)(X2 - Xi)
(by the Mean Value Theorem)
f(b) - f(x -,) = v(x,*)(b - x, _)
(by the Mean Value Theorem).
The left sides add to f(b) -f(a). The sum on the right, as Ax -- 0, is
J
v(x) dx.
APPLICATIONS OF INTEGRATION
Up to now the integral has been the area under a curve. There are many other
applications, quite different from areas. Whenever additionbecomes "continuous," we
have integralsinstead of sums. Chapter 8 has space to develop more applications, but
four examples can be given immediately--which will make the point.
We stay with geometric problems, rather than launching into physics or engineering
or biology or economics. All those will come. The goal here is to take a first step
away from rectangles.
EXAMPLE 3 (for circles) The area A and circumference C are related by dA/dr = C.
The question is why. The area is 7r2. Its derivative 27nr is the circumference. By the
Fundamental Theorem, the integral of C is A. What is missing is the geometrical
reason. Certainly rr2 is the integral of 2nrr, but what is the real explanation for A =
J C(r) dr?
My point is that the pieces are not rectangles. We could squeeze rectangles under
a circular curve, but their heights would have nothing to do with C. Our intuition
has to take a completely different direction, and add up the thin rings in Figure 5.16.
shell volume = 4ntr 2Ar
Fig. 5.16 Area of circle = integral over rings. Volume of sphere = integral over shells.
Suppose the ring thickness is Ar. Then the ring area is close to C times Ar. This is
precisely the kind of approximation we need, because its error is of higher order (Ar)2.
The integral adds ring areasjust as it added rectangular areas:
A=
C dr =
2nr dr =
Ir 2 .
That is our first step toward freedom, away from rectangles to rings.
218
5 Integrals
The ring area AA can be checked exactly-it is the difference of circles:
AA = ir(r + Ar) 2 - trr2 = 2rr Ar + 7r(Ar)2 .
This is CAr plus a correction. Dividing both sides by Ar - 0 leaves dA/dr = C.
Finally there is a geometrical reason. The ring unwinds into a thin strip. Its width
is Ar and its length is close to C. The inside and outside circles have different perimeters, so this is not a true rectangle-but the area is near CAr.
EXAMPLE 4
For a sphere, surface area and volume satisfy A = dV/dr.
What worked for circles will work for spheres. The thin rings become thin shells. A
shell goes from radius r to radius r + Ar, so its thickness is Ar. We want the volume
of the shell, but we don't need it exactly. The surface area is 47rr 2 , so the volume is
about 47rr 2 Ar. That is close enough!
Again we are correct except for (Ar)2 . Infinitesimally speaking dV= A dr:
V=
A dr =
4rr2 dr = rr3 .
This is the volume of a sphere. The derivative of V is A, and the shells explain why.
Main point: Integration is not restrictedto rectangles.
EXAMPLE 5 The distance around a square is 4s. Why does the area have dAlds = 2s?
The side is s and the area is s2. Its derivative 2s goes only half way aroundthe square.
I tried to understand that by drawing a figure. Normally this works, but in the figure
dAlds looks like 4s. Something is wrong. The bell is ringing so I leave this as an
exercise.
EXAMPLE6
Find the area under v(x)= cos-
x from x= 0 to x= 1.
That is a conventional problem, but we have no antiderivative for cos- x. We could
look harder, and find one. However there is another solution-unconventional but
correct. The region can be filled with horizontal rectangles (not vertical rectangles).
Figure 5.17b shows a typical strip of length x = cos v (the curve has v = cos'-x). As
the thickness Av approaches zero, the total area becomes x dv. We are integrating
J
upward, so the limits are on v not on x:
area
= O2
cos v dv = sin v]-'
2
=
1.
The exercises ask you to set up other integrals-not always with rectangles. Archimedes used triangles instead of rings to find the area of a circle.
------
t
OS-lX
S
OS V
s
s
do
S
T
AA = 4sAs?
dx 1
Fig. 5.17 Trouble with a square. Success with horizontal strips and triangles.
219
5.7 The Fundamental Theorem and its Applications
5.7
EXERCISES
Read-through questions
The area f(x) = J v(t) dt is a function of a . By Part 1 of
the Fundamental Theorem, its derivative is b . In the
proof, a small change Ax produces the area of a thin c
This area Af is approximately d times o . So the
derivative of J t 2 dt is
f
The integral Sb t 2 dt has derivative
. The minus sign
.
is because h . When both limits a(x) and b(x) depend on
x, the formula for df/dx becomes I minus __j_.In the
example
t dt, the derivative is
X"
k
By Part 2 of the Fundamental Theorem, the integral of
df/dx is I . In the special case when df/dx = 0, this says
that m . From this special case we conclude: If dA/dx =
dB/dx then A(x) = n . If an antiderivative of 1/x is Inx
(whatever that is), then automatically 1Sbdx/x = o
The square 0< x < s, 0 < y < s has area A = p. If s
is increased by As, the extra area has the shape of
.....
That area AA is approximately r . So dA/ds = s
Find the derivatives of the following functions F(x).
2
xfCoS t dt
2 1Scos 3t dt
2
4 JSx"dt
1
t" dt
2
fX U du
3
6 Sfx v(u) du
7 jx+1 v(t) dt (a "running average" of v)
1
tX
v(t) dt (the average of v; use product rule)
8 -
x
o
sin 2 t dt
9 -
2
So[fo v(u) du] dt
Jo v(t) dt + Sl v(t) dt
sin
t
2
dt
17 Sx u(t)v(t) dt
19
10 1 0x
o
fXX
25 If
JSx v(t)
dt = Sx v(t) dt (equal areas left and right of
function. Take derivatives to
zero), then v(x) is an
prove it.
26 Example 2 said that 2x dt/t does not really depend on x
(or t!). Substitute xu for t and watch the limits on u.
27 True or false, with reason:
(a) All continuous functions have derivatives.
(b) All continuous functions have antiderivatives.
(c) All antiderivatives have derivatives.
(d) A(x) = J~ dt/t 2 has dA/dx = 0.
Find
f~
v(t) dt from the facts in 28-29.
28 dx = v(x)
29
o v(t) dt-
o
x
x+2"
30 What is wrong with Figure 5.17? It seems to show that
dA = 4s ds, which would mean A = J 4s ds = 2s2.
31 The cube 0 < x, y, z s has volume V=
. The
three square faces with x = s or y = s or z = s have total area
A =
. If s is increased by As, the extra volume has
the shape of
. That volume AV is approximately
. So dV/ds =
32 The four-dimensional cube 0 < x, y, z, t < s has hypervolume H=
. The face with x= s is really a
. The total volume of
. Its volume is V =
the four faces with x = s, y = s, z = s, or t = s is
When s is increased by As, the extra hypervolume is
x +2
x
24 Suppose df/dx = 2x. We know that d(x 2)/dx = 2x. How
do we prove that f(x) = x 2 + C?
sin x sin- t dt
0o
t 3 dt
x
12 jx
(df/dt)2
dt
14 Sx v(- t) dt
16 Ix sin t dt
18 J(x) 5 dt
f(x)
20
x)dfdt
dt
21 True or false
If df/dx = dg/dx then f(x) = g(x).
If d2 f/dx2 = d2 g/dx2 then f(x) = g(x) + C.
If 3 > x then the derivative of fJ v(t) dt is - v(x).
The derivative of J1 v(x) dx is zero.
22 For F(x) = 1Sx sin t dt, locate F(n + Ax) - F(Xi) on a sine
graph. Where is F(Ax)- F(0)?
23 Find the function v(x) whose average value between 0 and
x is cos x. Start from fo v(t) dt = x cos x.
AH ;
. So dH/ds =
33 The hypervolume of a four-dimensional sphere is H =
2 4
-1
r . Therefore the area (volume?) of its three-dimensional
surface x 2 +y2 + Z2 + t2 = r2 is_
34 The area above the parabola y = x 2 from x = 0 to x = 1
is 4. Draw a figure with horizontal strips and integrate.
35 The wedge in Figure (a) has area ½r2 dO. One reason: It is
a fraction dO/2n of the total area ,7r2. Another reason: It is
close to a triangle with small base rdO and height
Integrating ½r2 dO from 0 = 0 to 0 =
gives the area
of a quarter-circle.
2
36 A = So
- x
dx is also the area of a quarter-circle.
Show why, with a graph and thin rectangles. Calculate this
integral by substituting x = r sin 0 and dx = r cos 0 dO.
(c)
(b)
Sr
x
5 Integrals
37 The distance r in Figure (b) is related to 0 by r =
Therefore the area of the thin triangle is i r 2d0 =
Integration to 0 =
gives the total area 4.
38 The x and y coordinates in Figure (c) add to
. Without integrating explain why
41 The length of the strip in Figure (e) is approximately
. The width is
. Therefore the triangle has
area
da (do you get i?).
r cos 0 + r sin 0 =
42 The area of the ellipse in Figure (f) is 2zr2. Its derivative
is 4zr. But this is not the correct perimeter. Where does the
usual reasoning go wrong?
39 The horizontal strip at height y in Figure (d) has width dy
and length x =
. So the area up to y = 2 is
.
What length are the vertical strips that give the same area?
43 The derivative of the integral of v(x) is ~ ( x )What
.
is the
corresponding statement for sums and differencesof the numbers vj? Prove that statement.
40 Use thin rings to find the area between the circles r = 2
and r = 3. Draw a picture to show why thin rectangles would
be extra difficult.
44 The integral of the derivative of f (x) is f (x) + C. What is
the corresponding statement for sums of differences of f,?
Prove that statement.
45 Does d2f /dx2 = a(x)lead to
(It a(t) dt) dx =f ( I ) -f(O)?
46 The mountain y = - x2 + t has an area A(t) above the x
axis. As t increases so does the area. Draw an xy graph of the
mountain at t = 1. What line gives dA/dt? Show with words
or derivatives that d 2 ~ / d t>20.
5.8 Numerical Integration
This section concentrates on definite integrals. The inputs are y ( x ) and two endpoints
a and b. The output is the integral I. Our goal is to find that number
1; y(x) d x = I, accurately and in a short time. Normally this goal is achievable-as
soon as we have a good method for computing integrals.
Our two approaches so far have weaknesses. The search for an antiderivative
succeeds in important cases, and Chapter 7 extends that range-but generally f ( x )
is not available. The other approach (by rectangles) is in the right direction but too
crude. The height is set by y(x) at the right and left end of each small interval. The
right and left rectangle rules add the areas ( A x times y):
R,=(Ax)(y,+y,+
-.. + y , ) and L n = ( A x ) ( y o + y l + . - -+y,-,).
The value of y(x) at the end of interval j is yj. The extreme left value yo = y(a) enters
L, . With n equal intervals of length A x = ( b - a)/n, the extreme right value is y, =
y(b). It enters R,. Otherwise the sums are the same-simple to compute, easy to
visualize, but very inaccurate.
This section goes from slow methods (rectangles) to better methods (trapezoidal
and midpoint) to good methods (Simpson and Gauss). Each improvement cuts down
the error. You could discover the formulas without the book, by integrating x and
221
5.8 Numerical Integration
x 2 and x 4 . The rule R, would come out on one side of the answer, and L, would be
on the other side. You would figure out what to do next, to come closer to the exact
integral. The book can emphasize one key point:
The quality of a formula depends on how many integrals
1dx, f x dx, f x 2 dx, ..., it computes exactly. If f x Pdx
is the first to be wrong, the order of accuracy is p.
f
By testing the integrals of 1, x, x 2, ..., we decide how accurate the formulas are.
Figure 5.18 shows the rectangle rules R, and L,. They are already wrong when
y = x. These methods are first-order: p = 1. The errors involve the first power of
Ax-where we would much prefer a higher power. A larger p in (Ax) P means a
smaller error.
n
Yn
E=
E= -• Ax(yj+-1 -2
yj)
e=-E
··
Yn- 1
Yj+1
Yj+
¥
Y1
U
I
I
Ii
Ax
Ax
Ax
Ax
Yj
I
Ax
Ax
Fig. 5.18 Errors E and e in R. and L, are the areas of triangles.
When the graph of y(x) is a straight line, the integral I is known. The error triangles
E and e have base Ax. Their heights are the differences yj+ 1 - yj. The areas are
'(base)(height), and the only difference is a minus sign. (L is too low, so the error
L - I is negative.) The total error in R. is the sum of the E's:
R, - I = ½Ax(y - Yo) + -.- + ½Ax(y - yn-_1)=
Ax(y. - yo).
(1)
All y's between Yo and y, cancel. Similarly for the sum of the e's:
L- - I
- ½Ax(Yn - Yo)
=
-
Ax[y(b - y(a)].
(2)
The greater the slope of y(x), the greater the error-since rectangles have zero slope.
Formulas (1) and (2) are nice-but those errors are large. To integrate y = x from
a = 0 to b = 1, the error is ½Ax(1 - 0). It takes 500,000 rectangles to reduce this error
to 1/1,000,000. This accuracy is reasonable, but that many rectangles is unacceptable.
The beauty of the error formulas is that they are "asymptotically correct" for all
functions. When the graph is curved, the errors don't fit exactly into triangles. But
the ratio of predicted error to actual error approaches 1. As Ax -+ 0, the graph is
almost straight in each interval-this is linear approximation.
The error prediction ½Ax[y(b)- y(a)] is so simple that we test it on y(x) = x:
I=
o
dx
1
10
100
1000
error R - I=
.33
.044
.0048
.00049
error L, - I=
-. 67
-. 056
-. 0052
-. 00051
n
The error decreases along each row. So does Ax = .1, .01, .001, .0001. Multiplying n
by 10 divides Ax by 10. The error is also divided by 10 (almost). The error is nearly
proportional to Ax-typical of first-order methods.
The predicted error is ½Ax, since here y(1) = 1 and y(O) = 0. The computed errors
in the table come closer and closer to ½Ax = .5, .05, .005, .0005. The prediction is the
"leading term" in the actual error.
222
5
Integrals
The table also shows a curious fact. Subtracting the last row from the row above
gives exact numbers 1, .1, .01, and .001. This is (R, - I) - (L, - I), which is R, - L,.
It comes from an extra rectangle at the right, included in R. but not L,. Its height is
1 and its area is 1, .1, .01, .001.
The errors in R. and L. almost cancel. The average T, = ½(R, + L,) has less errorit is the "trapezoidal rule." First we give the rectangle prediction two final tests:
n= l
errors
J (x 2 - x) dx:
J dx/(l0 + cos 2nx): errors
n= 10
n = 100
3
1.7 10- '
1.7 10-
10-3
2 . 10-'4
-1
n= 1000
1.7*10 -
1.7 10-5
7
"0"
"0"
2
Those errors are falling faster than Ax. For y = x - x the prediction explains why:
y(O) equals y(l). The leading term, with y(b) minus y(a), is zero. The exact errors are
'(Ax) 2 , dropping from 10-1 to 10- 3 to 10- 5 to 10- 7 . In these examples L, is identical
to R. (and also to T,), because the end rectangles are the same. We will see these
((Ax) 2 errors in the trapezoidal rule.
The last row in the table is more unusual. It shows practically no error. Why do
the rectangle rules suddenly achieve such an outstanding success?
The reason is that y(x) = 1/(10 + cos 2nrx) is periodic. The leading term in the error
is zero, because y(O) = y(l). Also the next term will be zero, because y'(0) = y'(1). Every
power of Ax is multiplied by zero, when we integrate over a complete period. So the
errors go to zero exponentially fast.
Personalnote I tried to integrate 1/(10 + cos 27rx) by hand and failed. Then I was
embarrassed to discover the answer in my book on applied mathematics. The method
was a special trick using complex numbers, which applies over an exact period.
Finally I found the antiderivative (quite complicated) in a handbook of integrals, and
verified the area 1/-99.
THE TRAPEZOIDAL AND MIDPOINT RULES
We move to integration formulas that are exact when y = x. They have secondorder accuracy. The Ax error term disappears. The formulas give the correct area
under straight lines. The predicted error is a multiple of (Ax) 2. That multiple is found
by testing y = x 2 -for which the answers are not exact.
The first formula combines R. and L,. To cancel as much error as possible, take
the average !(R, + L,). This yields the trapezoidal rule, which approximates
Sy(x) dx by Tn:
(3)
+ ULn= Ax(½yo + yl + Y2 + .. + y.n-1 + yn).
RT.=
Another way to find T.is from the area of the "trapezoid" below y = x in Figure 5.19a.
Tn =-Ax
I(Yo + )++ - I(Y1 + Y2)
2 2
+
"'
Yn
1
E=
12
(Ax)2 (
V,,
)
"J+
1J
e=--I E
2
yi
YO
I
Ax
Fig. 5.19
Ax
Ax
Ax
j+l
j
Ax
Second-order accuracy: The error prediction is based on v = x 2.
j+ 1
5.8 Numerical Intqrdon
The base is Ax and the sides have heights yj-l and yj. Adding those areas gives
+(L, + R,) in formula (3)-the coefficients of yj combine into f + f = 1. Only the first
and last intervals are missing a neighbor, so the rule has f yo and f y,. Because
trapezoids (unlike rectangles) fit under a sloping line, T,, is exact when y = x.
What is the difference from rectangles? The trapezoidal rule gives weight f Ax to
yo and y,. The rectangle rule R, gives full weight Ax to y, (and no weight to yo).
R, - T, is exactly the leading error f y, - +yo.The change to T,, knocks out that error.
Another important formula is exact for y(x) = x. A rectangle has the same area as
a trapezoid, if the height of the rectangle is halfway between yj- and yj . On a straight
line graph that is achieved at the midpoint of the interval. By evaluating y(x) at the
halfway points f Ax, AX, AX, ..., we get much better rectangles. This leads to the
midpoint rule M n:
For
1; x dx, trapezoids give f (0)+ 1 + 2 + 3 + f(4) = 8. The midpoint rule gives
4 + 4 +3 + 3 = 8, again correct. The rules become different when y = x2, because y,,,
is no longer the average of yo and y,. Try both second-order rules on x2:
I=
x2 dx
1
10
100
error T, - I =
116
l/600
1/60000
error M , - I =
-1112
-1/1200
-1/120000
n=
The errors fall by 100 when n is multiplied by 10. The midpoint rule is twice as good
(- 1/12 vs. 116). Since all smooth functions are close to parabolas (quadratic approximation in each interval), the leading errors come from Figure 5.19. The trapezoidal
error is exactly
when y(x) is x2 (the 12 in the formula divides the 2 in y'):
For exact error formulas, change yt(b)- yt(a) to (b - a)yM(c).The location of c is
unknown (as in the Mean Value Theorem). In practice these formulas are not much
used-they involve the pth derivative at an unknown location c. The main point
about the error is the factor AX)^.
One crucial fact is easy to overlook in our tests. Each value of y(x) can be extremely
hard to compute. Every time a formula asks for yj, a computer calls a subroutine. The
goal of numerical integration is to get below the error tolerance, while calling for a
minimum number of values of y. Second-order rules need about a thousand values for
a typical tolerance of
The next methods are better.
FOURTH-ORDER RULE: SIMPSON
The trapezoidal error is nearly twice the midpoint error (116 vs. - 1/12). So a
good combination will have twice as much of M, as T,. That is Simpson's rule:
Multiply the midpoint values by 213 = 416. The endpoint values are multiplied by
224
5 Integrals
2/6, except at the far ends a and b (with heights Yo and y,). This 1-4-2-4-2-4-1
pattern has become famous.
Simpson's rule goes deeper than a combination of T and M. It comes from a
parabolic approximation to y(x) in each interval. When a parabola goes through yo,
Yl/2, yl, the area under it is !Ax(yo + 4 yl/2+ YI). This is S over the first interval. All
our rules are constructed this way: Integrate correctly as many powers 1, x, x 2,
...
as
possible. Parabolas are better than straight lines, which are better than flat pieces.
S beats M, which beats R. Check Simpson's rule on powers of x, with Ax = 1/n:
n= 1
n= 10
n= 100
y= x2
0
0
0
error if y = x3
0
0
0
8.33 - 10-3
8.33 10-7
8.33.10-11
error if
error if y =
x4
Exact answers for x 2 are no surprise. S, was selected to get parabolas right. But the
zero errors for x3 were not expected. The accuracy has jumped to fourth order, with
errors proportional to (Ax)4 . That explains the popularity of Simpson's rule.
To understand why x3 is integrated exactly, look at the interval [-1, 1]. The odd
function x3 has zero integral, and Simpson agrees by symmetry:
Sx
dx =
x
3
=0
and
4
6
2
yn
4
(8)
13 =0.
[(-1)3 +4(0)3+
1
1
6
2
2
Y(
G
Fig. 5.20
Ax
j
j+1I
.I
Ax
Ax
Ax
j+1
Ax/f-
Simpson versus Gauss: E = c(Ax)4 (yj'i 1 - yj") with cs = 1/2880 and c, = - 1/4320.
THE GAUSS RULE (OPTIONAL)
We need a competitor for Simpson, and Gauss can compete with anybody. He
calculated integrals in astronomy, and discovered that two points are enough for a
fourth-order method. From -1 to 1 (a single interval) his rule is
I_ y(x) dx ?%y(- 1//3) + y(1/-,3).
(9)
Those "Gauss points" x = - 1/,3 and x = 1/,3 can be found directly. By placing
them symmetrically, all odd powers x, x 3, ... are correctly integrated. The key is in
y = x 2 , whose integral is 2/3. The Gauss points - x, and + XG get this integral right:
2
1
- (- xG)2
3
(X
)2 ,
SO x
=
3
and
x,
= +
1
Figure 5.20c shifts to the interval from 0 to Ax. The Gauss points are
(1 ± 1/ •) Ax/2. They are not as convenient as Simpson's (which hand calculators
prefer). Gauss is good for thousands of integrations over one interval. Simpson is
5.8
Numerical Integration
good when intervals go back to back-then Simpson also uses two y's per interval.
For y = x4, you see both errors drop by l o p 4 in comparing n = I to n = 10:
I = 1; x4 dx
Simpson error
Gauss error
8.33 l o p 3
- 5.56
8.33 l o p 7
- 5.56
lop7
DEFINITE INTEGRALS ON A CALCULATOR
It is fascinating to know how numerical integration is actually done. The points are
not equally spaced! For an integral from 0 to 1, Hewlett-Packard machines might
internally replace x by 3u2 - 2u3 (the limits on u are also 0 and 1). The machine
remembers to change dx. For example,
:1 5
becomes
Algebraically that looks worse-but the infinite value of l/& at x = 0 disappears
at u = 0. The differential 6(u - u2) du was chosen to vanish at u = 0 and u = 1. We
don't need y(x) at the endpoints-where infinity is most common. In the u variable
the integration points are equally spaced-therefore in x they are not.
When a difficult point is inside [a, b], break the interval in two pieces. And chop
off integrals that go out to infinity. The integral of e p x 2should be stopped by
x = 10, since the tail is so thin. (It is bad to go too far.) Rapid oscillations are among
the toughest- the answer depends on cancellation of highs and lows, and the calculator requires many integration points.
The change from x to u affects periodic functions. I thought equal spacing was
good, since 1/(10+ cos 2nx) was integrated above to enormous accuracy. But there
is a danger called aliasing. If sin 8nx is sampled with Ax = 118, it is always zero. A
high frequency 8 is confused with a low frequency 0 (its "alias" which agrees at the
sample points). With unequal spacing the problem disappears. Notice how any integration method can be deceived:
Ask for the integral of y = 0 and specify the accuracy. The calculator
samples y at x,, . . ., x,. (With a PAUSE key, the x's may be displayed.)
Then integrate Y(x) = (x - x , ) ~ (x - x , ) ~ . That also returns the
answer zero (now wrong), because the calculator follows the same steps.
On the HP-28s you enter the function, the endpoints, and the accuracy. The
variable x can be named or not (see the margin). The outputs 4.67077 and 4.7E-5 are
the requested integral ex dx and the estimated error bound. Your input accuracy
.00001 guarantees
3: ( ( E X P I )
3: ' E X P ( X 1 '
true y - computed y
relative error in y =
<
.00001.
2: € 1 23
2 : € X 1 2)
computed y
1 : .00001
1 : .00001
The machine estimates accuracy based on its experience in sampling y(x). If you
guarantee ex within .00000000001, it thinks you want high accuracy and takes longer.
In consulting for HP, William Kahan chose formulas using 1, 3, 7, 15, ... sample
points. Each new formula uses the samples in the previous formula. The calculator
stops when answers are close. The last paragraphs are based on Kahan's work.
5 Integrals
TI-81 Program to Test the Integration Methods L, R, T, M , S
Prgm1:NUM I N T
: D i s p "A="
:Input A
:D iS P IIB=~I
:Input B
:Lbl N
: D i s p "N="
:Input N
: ( B - A ) /N+D
:D/2+H
:A+X
: Y p L
:l+J
:@+R
:8 + M
:LbL I
:X+H+X
: M + Y l -+M
:A+JD-,X
:R+Yl+R
:IS>(J,N)
:Goto 1
:(L+R-Yl)D+L
:R D + R
:MD+M
: ( L + R ) /2+T
:( 2 M t T ) / 3 + S
:Disp
T,
:Disp
:Disp
:Disp
:Disp
:Disp
:Pause
"L,
S"
L
R
M
T
S
R,
M,
:Goto N
Place the integrand y(x) in the Y 1 position on the Y = function edit screen. Execute
this program, indicating the interval [A, B ] and the number of subintervals N. Rules
L and R and M use N evaluations of y(x). The trapezoidal rule uses N + 1 and
Simpson's rule uses 2N + 1. The program pauses to display the results. Press ENTER
to continue by choosing a different N. The program never terminates (only pauses).
You break out by pressing ON. Don't forget that IS , G o t o, ... are on menus.
5.8 EXERCISES
Read-through questions
To integrate y(x), divide [a, b] into n pieces of length
b
over each piece,
Ax = a . R, and L, place a
using the height at the right or
c
endpoint:
+ y,) and L, = d . These are e
R, = Ax(yl +
order methods, because they are incorrect for y = f . The
total error on [0,1] is approximately Q . For y = cos ax
this leading term is h . For y = cos 2nx the error is very
small because [0, 1) is a complete i .
+
A much better method is T,=$Rn
i =
Ax[iyo + k y1 +
+ L y , ] . This m rule is
n -order because the error for y = x is
o . The error
rule is twice as
for y = x2 from a to b is P . The CI
accurate, using M, = Ax[ r 1.
Simpson's method is S, = $Mn+ s . It is t -order,
u
are integrated correctly. The
because the powers
times Ax. Over three
coefficients of yo, yIl2,yl are v
intervals the weights are Ax16 times 1-4- w . Gauss uses
x
points in each interval, separated by ~ x / f i For a
method of order p the error is nearly proportional to Y .
1 What is the difference L, - T,? Compare with the leading
error term in (2).
2 If you cut Ax in half, by what factor is the trapezoidal
error reduced (approximately)?By what factor is the error in
Simpson's rule reduced?
3 Compute Rn and Ln for x3dx and n = l,2,10. Either
verify (with computer) or use (without computer) the formula
l 3 + 23 +
+ n3 = tn2(n+
4 One way to compute T,, is by averaging i(L, + R,).
Another way is to add iyo + yl +
+ iy,. Which is more
efficient? Compare the number of operations.
5 Test three different rules on I =
6 Compute n to six places as 4
rule.
x4dx for n = 2 4 , 8.
1; dx/(l + x2), using any
+
7 Change Simpson's rule to Ax($yo 4 yllz
interval and find the order of accuracy p.
+ 4y ) in each
8 Demonstrate superdecay of the error when 1/(3+ sin x) is
integrated from 0 to 2a.
9 Check that ( A ~ ) ~ ( y j +yj)/12
,
is the correct error for
y = 1 and y = x and y = x2 from the first trapezoid ( j= 0).
Then it is correct for every parabola over every interval.
10 Repeat
Problem 9 for the midpoint error
-(A~)~(yj+ yj)/24. Draw a figure to show why the rectangle M has the same area as any trapezoid through the midpoint (including the trapezoid tangent to y(x)).
11 In principle
sin2x dx/x2 = n. With a symbolic algebra code or an HP-28S, how many decimal places do you
get? Cut off the integral to I,!
and test large and small A.
12 These four integrals all equal n:
LJ& I-rn
m
=dx
x
1'-
- 112 dx
l+x
(a) Apply the midpoint rule to two of them until
n x 3.1416.
(b) Optional: Pick the other two and find a x 3.
5.8 Numerical Intogrotion
13 To compute in 2 = dx/x = .69315 with error less than
.001, how many intervals should T, need? Its leading error is
AX)^ [yt(b)- yt(a)]/12. Test the actual error with y = llx.
I; &
14 Compare T. with M nfor
dx and n = 1,10,100. The
error prediction breaks down because yt(0)= oo.
1;
15 Take f(x) = y(x) dx in error formula 3R to prove that
y(x) dx - y(0)Ax is exactly f (AX)~Y'(C)
for some point c.
16 For the periodic function y(x) = 1/(2+ cos 6zx) from -1
to 1, compare T and S and G for n = 2.
17 For I =
dal rule is
1;
dx, the leading error in the trapezoi. Try n = 2,4,8 to defy the prediction.
18 Change to x = sin 8, ,/= cos 8, dx = cos 8 dB, and
repeat T, on j;l2 cos28 dB. What is the predicted error after
the change to O?
19 Write down the three equations Ay(0)+ By($) + Cy(1)= I
for the three integrals I = 1 dx, x dx, x2 dx. Solve for
A, B, C and name the rule.
1;
:I
1;
20 Can you invent a rule using Ay, + Byll4 + CyIl2+
Dy3/, Ey, to reach higher accuracy than Simpson's?
+
21 Show that T, is the only combination of L, and R, that
has second-order accuracy.
22 Calculate 1e-x2 dx with ten intervals from 0 to 5 and 0
What
to 20 and 0 to 400. The integral from 0 to m is f
is the best point to chop off the infinite integral?
&.
+
23 The graph of y(x) = 1/(x2 10- l o ) has a sharp spike and
a long tail. Estimate y dx from Tlo and Tloo(don't expect
sec28 d0
much). Then substitute x = 10- tan 8, dx =
and integrate lo5 from 0 to 44.
1;
24 Compute Jx- nl dx from T, and compare with the
divide and conquer method of separating lx - n( dx from
Ix - nl dx.
1;
25 Find a, b, c so that y = ax2 + bx + c equals 1,3,7 at
x = 0, 3, 1 (three equations). Check that 4 1 8 3 4 7
equals y dx.
1;
+
+
26 Find c in S - I = AX)^ [yftt(l)- yt"(0)] by taking y = x4
and Ax = 1.
27 Find c in G - I = ~(Ax)~[y"'(l)
- y"'(- 1)] by taking
y = x4, Ax = 2, and G = (- l ~ f l )+~(l/fi14.
28 What condition on y(x) makes L,= R, = T, for the
integral y(x) dx?
29 Suppose y(x) is concave up. Show from a picture that the
trapezoidal answer is too high and the midpoint answer is
too low. How does y" > 0 make equation (5) positive and (6)
negative?
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Resource: Calculus Online Textbook
Gilbert Strang
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